International Journal of Algebra, Vol. 13, 2019, no. 8, 407 - 430 HIKARI Ltd, www.m-hikari.com https://doi.org/10.12988/ija.2019.91037 On Some Diophantine Equations of the Form hX 4 + Y 4 = Z n GustafS¨oderlund Kettilsgatan 4A 582 21 Link¨oping, Sweden This article is distributed under the Creative Commons by-nc-nd Attribution License. Copyright c 2019 Hikari Ltd. Abstract We show the impossibility of primitive non-zero solutions to the equation hx 4 + y 4 = z n if h = 43 and n ∈{2, 3, 4, 5, 6} and if h ∈{3, 19, 67} and n ∈{3, 4, 5, 6} and finally if h ∈{11, 163} and n ∈{4, 5}. The proofs are based mostly on elementary methods with no use of computers and the elliptic curve machinery with the ex- ception of a few cases where the rank determinations of certain elliptic curves were accomplished. Mathematics Subject Classification: 11D41 Keywords: Diophantine equations, generalized Fermat equations, primi- tive non-zero solutions 1 Introduction The title equation belongs to a group of diophantine equations called ”generalized Fermat equations” which in its general form can be written as Ax p + By q = Cz r where p, q and r are fixed integers > 1 and A, B and C are fixed non-zero integers [8]. According to a theorem of Darmon and Granville this equation has only a finite number of primitive non-zero solutions (i.e. Ax, By and Cz are pairwise relatively prime and x · y · z 6= 0) if 1 p + 1 q + 1 r < 1 [8]. This means that under these conditions the equation has no solution at all or a finite number of non-zero solutions. Hence we conclude that the title equa- tion has no solution or a finite number of non-zero solutions if n> 2. Recently the author of this paper proved that the diophantine equation 2x 4 +y 4 = z 3 has
Transcript
International Journal of Algebra, Vol. 13, 2019, no. 8, 407 - 430HIKARI Ltd, www.m-hikari.com
https://doi.org/10.12988/ija.2019.91037
On Some Diophantine Equations of the Form
hX4 + Y 4 = Zn
Gustaf Soderlund
Kettilsgatan 4A 582 21 Linkoping, Sweden
This article is distributed under the Creative Commons by-nc-nd Attribution License.
The title equation belongs to a group of diophantine equations called”generalized Fermat equations” which in its general form can be written asAxp + Byq = Czr where p, q and r are fixed integers > 1 and A,B and C arefixed non-zero integers [8]. According to a theorem of Darmon and Granvillethis equation has only a finite number of primitive non-zero solutions (i.e.Ax,By and Cz are pairwise relatively prime and x · y · z 6= 0) if 1
p+ 1
q+ 1
r< 1
[8]. This means that under these conditions the equation has no solution at allor a finite number of non-zero solutions. Hence we conclude that the title equa-tion has no solution or a finite number of non-zero solutions if n > 2. Recentlythe author of this paper proved that the diophantine equation 2x4+y4 = z3 has
408 Gustaf Soderlund
the only primitive non-zero solutions (x, y, z) = (±5,±3, 11) [18]. In the early1990s Terai and Osada and Cao published two papers concerning the similarequations x4 + dy4 = zp and cx4 + dy4 = zp, where p is an odd prime. Theyshowed that these equations have no integer solutions if certain conditionsare fulfilled [4,19]. However these results are not applicable on the equationspresented in this paper. The equation hx2 + y4 = zn has been examined forall n = 4 solely for h ∈ {1, 2}. Thus Ellenberg, using the method of Galoisrepresentations and modularity, proved the non-existence of primitive non-zerosolutions to the equation x2+y4 = zp for p a prime number = 211 [11]. Dieule-fait and Urroz using similar methods found no primitive non-zero solutions tothe equations 2x2 + y4 = zp and 3x2 + y4 = zp for any prime number p > 349and p > 131 respectively [9]. The results, when h ∈ {1, 2} were completed byBennett, Ellenberg and Ng showing that the equation x2+y4 = zn has no prim-itive non-zero solutions if n = 4 and that the only primitive positive non-zerosolution to the equation 2x2 + y4 = zn for all n = 4 is (x, y, z, n) = (11, 1, 3, 5)[1]. However most publications concerning the generalized fermat equationdeal with the case when A = B = C = 1 [2,3,15] which is associated with aconjecture stating that the equation xp + yq = zr has no primitve non-zerosolutions if min (p, q, r) ≥ 3 [13]. Finally, should the abc-conjecture become atheorem there are only a finite number of primitive non-zero solutions to theequation Axp + Byq = Czr for A,B,C fixed nonzero integers and all posi-tive p, q, r such that 1
p+ 1
q+ 1
r< 1, where solutions arising from the identity
1p + 23 = 32 are regarded as just one [12].
2 Main results
Theorem 2.1 The diophantine equation 43x4 + y4 = z2 has no non-zerosolutions.
Proof. According to H. Cohen this equation has non-zero solutions if andonly if the corresponding elliptic curve Y2 = X3 + 43X has non-zero rank [5]and from J. Cremonas database [6] we find that this curve has indeed rankzero.
Theorem 2.2 The diophantine equation 3x4 + y4 = z3 has no primitivenon-zero solutions.
Proof. We thus assume that 3x, y and z are pairwise relatively prime andx · y · z 6= 0. Hence we must after congruence considerations conclude thatx 6≡ ymod 2 and z is odd. We have,
3(x2)2 + (y2)2 = z3
On some Diophantine equations ... 409
However according to [10] ∃ integers a and b such that y2 + x2√−3 =
(a+ b√−3)3. Hence,
y2 = a(a2 − 9b2) and x2 = 3b(a2 − b2). a 6≡ bmod 2since z is odd and (a, b) = 1 since (x, y) = 1. 3 - a since 3 - y and we have(a, a2 − 9b2) = 1.
Case I. 3 | b and 3 - aHence b = 3t =⇒ 3 - (a2 − b2) since (a, b) = 1. Thus (3b, a2 − b2) = 1 and
from x2 = 3b(a2 − b2) we see that b = ±3v2 (1). From y2 = a(a2 − 9b2) wehave a = ±V 2 (2) and a2−9b2 = ±U2
1 where the negative sign is rejected afterexamination modulo 3 of the last equation. Hence a2− 9b2 = U2
1 (3) and from(1), (2) and (3) we have V 4− (3v)4 = U2
1 and this well-known equation has nonon-zero solutions [16].
Case II. 3 - b and 3 - aThus 3u | (a2 − b2) and u must be odd since from x2 = 3b(a2 − b2) (1) we
see that x2 must be divided by an even power of 3.Hence equation (1) can be written x2 = 3u+1 · b · (a2−b2
3u) and (b, a
2−b23u
) = 1.
Thus, b = ±T 2 and a2−b23u
= ±U22 =⇒ a2− b2 = ±3 · (3u−1
2 ·U2)2 =⇒ a2−T 4 =
±3U23 (2). However from y2 = a(a2 − 9b2) and since the two factors on RHS
are coprime we see that a = ±V 2. Hence from (2) we have V 4 − T 4 = ±3U23
and this last equation has no non-zero solution according to [5] since| ±3 |= 3is not a congruent number.
Theorem 2.3 The diophantine equation 19x4 + y4 = z3 has no primitivenon-zero solutions.
Proof. From prerequisites we see that 19x, y and z are pairwise relativelyprime and x · y · z 6= 0 and after congruence considerations we realize thatx 6≡ ymod 2 and z is odd. We get,
Subcase 1. 3 - c and 3 - dWe have (c, c2− 57d2) = 1 and (d, 3c2− 19d2) = 1. Thus from equation (1)
we get,
c = ±V 21 (3). From equation (2) we have d = ±V 2
2 (4)and 3c2−19d2 = ±U2
2 (5) where the positive sign is rejected after examinationmodulo 3 of equation (5). From (4) and (5) we have 3c2 − 19V 4
2 = −U22 (6).
(3) + (6) =⇒ 3V 41 − 19V 4
2 = −U22 (7). If V1 is odd and V2 is even we have
19V 42 = 3V 4
1 + U22 , but LHS ≡ 0 mod 16 and RHS 6≡ 0 mod 16 and we have a
contradiction. If V1 is even and V2 is odd we have from (7), 3V 41 = 19V 4
2 − U22
and we have again a contradiction after congruence considerations.
Subcase 2. 3 | c and 3 - dThus c = 3t. From equation (1) we have,
3t(9t2 − 57d2) = y2
9t(3t2 − 19d2) = y2. However, (9t, 3t2 − 19d2) = 1and we have,
9t = ±V 23 (8) and 3t2 − 19d2 = ±U2
3 (9). Herewe must reject the positive sign after modulo 3 examination of equation (9).From equation (8) we have t = −v2 (10). From equation (2) we have,
3 . U3 is odd and v 6≡ V4 mod 2. Soif v is even and V4 is odd we have 3v4 = 19V 4
4 − U23 and V L ≡ 0 mod 16 and
HL 6≡ 0 mod 16 thus a contradiction. If v is odd and V4 even we get 19V 44 =
3v4 + U23 and we have again a contradiction after modulo 16 examination.
Subcase 3. 3 - c and 3 | dHence d = 3s. From equation (2) we have,
3s(3c2 − 19 · 9s2) = x2
On some Diophantine equations ... 411
9s(c2 − 57s2) = x2. Hence (9s, c2 − 57s2) = 1 and wemust conclude that 9s = ±V 2
5 =⇒ s = ±V 26 (12). From equation (1) we get,
c(c2 − 57d2) = y2. Since (c, c2 − 57d2) = 1 weconclude that c = ±V 2
7 (13) and c2−57d2 = ±U25 (14) where the negative sign
is rejected after examination modulo 3 of equation (14). Hence from (12), (13)and (14) we have since d = 3s, V 4
7 − 57 · (±3V 26 )2 = U2
5 =⇒ V 47 − 513V 4
6 = U25
(15) and according to [5] this equation has non-zero solutions if and only ifthe corresponding elliptic curve Y 2 = X3−513X has non-zero rank. However,from [6] we conclude that this curve has rank zero and there are no non-zerosolutions to equation (15).
Case II. a and b are odd.Hence
y2 + x2√−19 = a3−19b3
√−19+3a2b
√−19−57ab2
8and we have,
8y2 = a3 − 57ab2 = a(a2 − 57b2) (1) and 8x2 = 3a2b− 19b3 = b(3a2 − 19b2)(2). (a, b) = 1 since (x, y) = 1. 19 - a since 19 - y.
Subcase 1. 3 - a and 3 - bHence we have (a, a2 − 57b2) = 1 and (b, 3a2 − 19b2) = 1. Thus from
equation (1) we get,a = ±U2
1 (3) and a2−57b2 = ±8V 21 (4) and the positive sign is rejected after
examination modulo 3 of equation (4). From (3) and (4) we get 57b2−U41 = 8V 2
1
(5). From equation (2) we have b = ±U22 (6) and 3a2−19b2 = ± 8V 2
2 (7) wherethe negative sign is rejected after examination modulo 3 of equation (7). Thus,3a2 − 19U4
2 = 8V 22 (8). From (3), (8), (5) and (6) we have,
3U41 − 19U4
2 = 8V 22 (9) and 57U4
2 − U41 = 8V 2
1 (10) where U1 and U2 areodd.
Furthermore we see that 3U1, 19U2 and 8V2 are pairwise relatively primeand so are 57U2, U1 and 8V1. 5 - U2. Indeed suppose that 5 | U2. Thus fromequation (10) we have 5 | (U4
1 + 8V 21 ) and 5 - U1 · V1. Hence U4
1 = 5K + 1and V 2
1 = 5L± 1 and we have 5 | (1± 8) and this is impossible. Suppose that5 | U1. Hence from equation (9) we get 5 | (19U4
2 + 8V 22 ). Since 5 - U2 · V2
we have U42 = 5M + 1 and V 2
2 = 5N ± 1. Hence 5 | 19 ± 8 and this isimpossible. Thus 5 - U1. Suppose finally that 5 | V2. From equation (9) wehave 5 | (3U4
1 −19U42 ). Since 5 - U1 ·U2 we have U4
1 = 5K+1 and U42 = 5M+1.
Thus 5 | (3− 19) = −16 and this is absurd and we must have V 22 = 5α± 1. So
if U41 = 5K + 1, U4
2 = 5M + 1 and V 22 = 5α± 1 we conclude from equation (9)
that 8, 24 ≡ 0 mod 5 and this is manifestly impossible. Thus we have shownthat subcase 1 is impossible.
5 (18) and a2 − 57β2 = ±8V 25 (19) where the positive sign is rejected
after examination modulo 3 of equation (19). Hence from from (18) and (19)we have a2 − 57U4
5 = −8V 25 (20). From equation (1) we have,
a(a2 − 57 · (3β)2) = 8y2
a(a2 − 57 · (−3U25 )2) = 8y2
a(a2− 513U45 ) = 8y2 where (a, a2− 513U4
5 ) = 1. Hencea = ±U2
6 (21) and a2− 513U45 = ±8V 2
6 (22) where the positive sign is rejectedafter examination modulo 3 of equation (22). Thus from (21) and (22) we haveU46 − 513U4
5 = −8V 26 (23) and from (20) and (21) we have U4
6 − 57U45 = −8V 2
5
(24). U5 and U6 are odd and after conguence considerations we realize that V5is odd and V6 is even. If we add equation (23) and equation (24) we get,
2U46 − 570U4
5 = −8V 25 − 8V 2
6
On some Diophantine equations ... 413
285U45 − U4
6 = 4(V 25 + V 2
6 )
16p+ 284 = 4(V 25 + V 2
6 )
4p+ 71 = V 25 + V 2
6
4p−V 26 = 8γ+ 1− 71 = 8γ− 70 and this is impossible
since 4p− V 26 ≡ 0 mod 4 and 8γ − 70 6≡ 0 mod 4.
Theorem 2.4 The diophantine equation 43x4 + y4 = z3 has no primitivenon-zero solutions.
Proof. From prerequisites we see that 43x, y and z are pairwise relativelyprime and x · y · z 6= 0 and after congruence considerations we realize thatx 6≡ ymod 2 and z is odd. We have,
(y2 + x2√−43) · (y2 − x2
√−43) = z3 where
y2 +x2√−43 and y2−x2
√−43 are coprime in the ring of integers (OQ
√−43) of
the numberfield Q√−43. Since OQ
√−43 has unique factorisation and all units
are cubes we get,
y2 + x2√−43 = (a+b
√−43
2)3 where a and b are integers of the same parity.
Case I. a and b are even.Hence,
y2 + x2√−43 = (c+ d
√−43)3 where c 6≡ dmod 2 since z is
odd.
y2 +x2√−43 = c3−43d3
√−43 + 3c2d
√−43−129cd2. Hence,
y2 = c(c2 − 129d2) (1) and x2 = d(3c2 − 43d2) (2). (c, d) = 1since (x, y) = 1. 43 - c since 43 - y.
Subcase 1. 3 - c and 3 - dHence (c, c2− 129d2) = 1 and (d, 3c2− 43d2) = 1 and from equation (1) we
have,
c = ±V 21 (3) and from equation (2) we get d = ±V 2
2 (4)and 3c2 − 43d2 = ±U2
2 (5) and the positive sign is rejected after examinationmodulo 3 of equation (5). From (3), (4) and (5) we have 3V 4
1 − 43V 42 = −U2
2
where V1 6≡ V2 mod 2 and U2 is odd. So if V1 is odd and V2 is even we have,
43V 42 = 3V 4
1 + U22 and after congruence considerations
we easily realize that this is impossible. If V1 is even and V2 is odd we get3V 4
1 = 43V 42 −U2
2 and again this is impossible after congruence considerations.
414 Gustaf Soderlund
Subcase 2. 3 | c and 3 - dThus c = 3t and from equation (1) we have,
3t(9t2 − 129d2) = y2
9t(3t2 − 43d2) = y2. However, (9t, 3t2 − 43d2) = 1and we get,
9t = ±V 23 =⇒ t = ±v2 (6) 3t2 − 43d2 = ±U2
3 (7) where the positive signis rejected after examination modulo 3 of equation (7). Thus from equation(6) and (7) we have 3v4 − 43d2 = −U2
3 (8). Furthermore we conclude that(d, 3c2 − 43d2) = 1 and from equation (2) we have d = ±V 2
4 (9). Hence from(8) and (9) we get 3v4 − 43V 4
4 = −U23 where U3 is odd and v 6≡ V4 mod 2. So
if V4 is odd and v is even we have,
3v4 = 43V 44 − U2
3 and RHS 6≡ 0 mod 4 andLHS ≡ 0 mod 4 and a contradiction is established. If V4 is even and v is oddwe get 43V 4
4 = 3v4 + U23 and this is impossible since 3v4 + U2
3 6≡ 0 mod 8 and43V 4
4 ≡ 0 mod 8.
Subcase 3. 3 - c and 3 | dThus d = 3s and from equation (2) we have,
4 (11). Here we mustreject the negative sign after examination modulo 3 of equation (11) and from(10) and (11) we have c2 − 129V 4
6 = U24 (12). From equation (1) we see that
(c, c2 − 129d2) = 1 and we must conclude that c = ±V 27 (13). Hence from
(12) and (13) we have V 47 − 129V 4
6 = U24 (14) where V7 6≡ V6 mod 2 and U4
is odd. However, according to [5] equation (14) has non-zero solutions if andonly if the corresponding elliptic curve Y 2 = X3 − 129X has non-zero rank.Unfortunately this elliptic curve is not included in [6] since the conductor istoo large. However in compliance with H. Daghigh and S. Didari [7] this curvehas rank zero.
Case II. a and b are odd.Hence
y2 +x2√−43 = a3−43b3
√−43+3a2b
√−43−129ab2
8and we have,
8y2 = a(a2 − 129b2) (1) and 8x2 = b(3a2 − 43b2) (2) where (a, b) = 1since (x, y) = 1. 43 - a since 43 - y.
On some Diophantine equations ... 415
Subcase 1. 3 - a and 3 - b.Hence we get (a, a2−129b2) = 1 and (b, 3a2−43b2) = 1. Thus from equation
(1) we have a = ±U21 (3) and a2 − 129b2 = ±8V 2
1 (4) where the positive signis rejected after examination modulo 3 of equation (4). Hence from equation(3) and (4) we have U4
1 − 129b2 = −8V 21 (5). From equation (2) we get
b = ±U22 (6) and 3a2 − 43b2 = ±8V 2
2 (7) where the negative sign is rejectedafter examination modulo 3 of equation (7). Hence from equation (6) and (7)we have 3a2−43U4
2 = 8V 22 (8). Equation (5)+(6) =⇒ U4
1 −129U42 = −8V 2
1 (9)where U1 and U2 are odd and V1 is even. Equation (3)+(8) =⇒ 3U4
1 −43U42 =
8V 22 (10) where V2 must be odd. From (10) - 3·(9) we have,
−43U42 + 3 · 129U4
2 = 8V 22 + 24V 2
1
43U42 = V 2
2 + 3V 21
3V 21 = 43U4
2 −V 22 . However this is impossible
since LHS ≡ 0 mod 4 and RHS 6≡ 0 mod 4.
Subcase 2. 3 | a and 3 - bHence a = 3t and from equation (1) we have,
8y2 = 3t(9t2 − 129b2)
8y2 = 9t(3t2− 43b2) where (9t, 3t2− 43b2) = 1.Thus 9t = ± odd square =⇒ t = ±U2
3 (11) and 3t2−43b2 = ±8V 23 (12) and the
negative sign is rejected after examination modulo 3 of equation (12). Hence3t2 − 43b2 = 8V 2
3 (13). From equation (2) we have,
8x2 = b(3a2− 43b2) where (b, 3a2− 43b2) = 1.Hence b = ±U2
4 (14) and 3a2 − 43b2 = ±8V 24 and the negative sign is rejected
after examination modulo 3 of the last equation. Thus 3a2 − 43b2 = 8V 24 (15)
and from equation (11), (14), (13) and (15) we have since a = 3t,
3U43 − 43U4
4 = 8V 23 (16) and,
27U43 − 43U4
4= 8V 24 (17). U3 and U4 are
odd and after congruence considerations we realize that V3 must be odd andV4 must be even. (17) - (16) =⇒ 3U4
U4 = C ·D where (C,D) = 1 and 3 - C ·D since (V4, 3V3) = 1.
N.B. Concerning the expressions of V3 and V4 in i1.) - i2.) and ii1.) - ii2) wehave V3 = ±(.......) and V4 = ±(.......). This certainly means that V3 = +(.......)and V4 = +(.......) or V3 = −(.......) and V4 = −(.......). If solutions exist oneparametric solution of V3 and V4 in i1.) - i2.) must be equal to at leastone parametric solution of V3 and V4 in ii1.) - ii2) for some value (values) ofA,B,C and D. Thus we have the following possibilities,
i1.) = ii1.): V4 = ± (3A4+B4)2
= ±(43C4+D4
2) (20). Since A4, B4, C4 and D4
are positive the signs in (20) are not independent and we have 3A4 + B4 =43C4 + D4 =⇒ B4 − D4 = 43C4 − 3A4 and since A,B,C and D are all oddwe have 43C4 − 3A4 ≡ 0 mod 16 which is impossible.
i2.) = ii1:): V4 = ± (A4+3B4)2
= ±(43C4+D4
2) which must be impossible
according to i1.) = ii1.).
i1.) = ii2:): V4 = ± (3A4+B4)2
= ±(C4+43D4
2) which must be impossible
according to i1.) = ii1.).
i2.) = ii2.): V4 = ± (A4+3B4)2
= ±(C4+43D4
2) which must be impossible
according to i1.) = ii1.).
Thus we have shown that subcase 2 is impossible.
Subcase 3. 3 - a and 3 | bHence b = 3s and from equation (2) we have,
From (31) we have 129U46 = (V6+V5)· (V6−V5) where (V6+V5, V6−V5) = 1.
Thus we have the following alternatives,
ii1.) V6 + V5 = ±129G4 and V6 − V5 = ±H4 and 129 - H. Hence
V5 = ±(129G4−H4)2
) and V6 = ±(129G4+H4
2).
ii2.) V6 + V5 = ±G4 and V6 − V5 = ±129H4 and 129 - G. Hence
V5 = ±(G4−129H4)
2) and V6 = ±(G
4+129H4
2).
ii3.) V6 + V5 = ±3G4 and V6 − V5 = ±43H4. 3 - H and 43 - G. Hence
V5 = ±(3G4−43H4)
2) and V6 = ±(3G
4+43H4
2).
ii4.) V6 + V5 = ±43G4 and V6 − V5 = ±3H4. 3 - G and 43 - H. Hence
V5 = ±(43G4−3H4)2
) and V6 = ±(43G4+3H4
2).
G ·H = U6 and (G,H) = 1 since (V6 + V5, V6 − V5) = 1.N.B. Concerning the expressions of V5 and V6 in i1.) and ii1.) - ii4) we
have V5 = ±(.......) and V6 = ±(.......). This certainly means that V5 = +(.......)and V6 = +(.......) or V5 = −(.......) and V6 = −(.......). If solutions exist atleast one parametric solution of V5 and V6 in i1.) must be equal to at leastone parametric solution of V5 and V6 in ii1.) - ii4) for some value (values) ofE,F,G and H. Thus we have the following possibilities,
418 Gustaf Soderlund
i1.) = ii1.): V6 = ±(E4+F 4
2) = ±(129G
4+H4
2) (32). Since E4, F 4, G4 and H4
are positive the signs in (32) are not independent. Hence, E4+F 4 = 129G4+H4
(33). V5 = ±(E4−F 4
6) = ±(129G
4−H4)2
) (34). Since the signs in (32) are notindependent so are the signs in (34) and we get, E4 − F 4 = 3 · 129G4 − 3H4
(35). (33) + (35) =⇒ 2E4 = 4 · 129G4 − 2H4 =⇒ E4 + H4 = 2 · 3 · 43G4 =⇒E4 +H4 ≡ 0 mod 3 and this is impossible since 3 - H · E.
i1.) = ii2.): V6 = ±(E4+F 4
2) = ±(G
4+129H4
2) and according to i1.) = ii1.)
we have, E4 + F 4 = G4 + 129H4 (36). V5 = ±(E4−F 4
6) = ±(G
4−129H4)2
) andaccording to i1.) = ii1.) we get, E4 − F 4 = 3G4 − 3 · 129H4 (37). (36) + (37)=⇒ 2E4 = 4G4 − 2 · 129H4 =⇒ 3 · 43H4 = 2G4 − E4 and this is impossiblesince 3 - G · E.
i1.) = ii3.): V6 = ±(E4+F 4
2) = ±(3G
4+43H4
2) and according to to i1.) = ii1.)
we have, E4 + F 4 = 3G4 + 43H4 (38). V5 = ±(E4−F 4
6) = ±(3G
4−43H4)2
) andaccording to i1.) = ii1.) we get, E4 − F 4 = 9G4 − 3 · 43H4 (39). (38) + (39)=⇒ 2E4 = 12G4 − 2 · 43H4. Thus, 6G4 = E4 + 43H4 and this is impossiblesince E4 + 43H4 ≡ 0 mod 4.
i1.) = ii4.): V6 = ±(E4+F 4
2) = ±(43G
4+3H4
2) and again according to i1.) =
ii1.) we have, E4 + F 4 = 43G4 + 3H4 (40). V5 = ±(E4−F 4
6) = ±(43G
4−3H4)2
)and again according to i1.) = ii1.) we get, E4 − F 4 = 3 · 43G4 − 9H4 (41).(40) + (41) =⇒ 2E4 = 4 · 43G4 − 6H4 =⇒ E4 + 3H4 = 2 · 43G4 and this isimpossible since E4 + 3H4 ≡ 0 mod 4.
This completes the proof of subcase 3 and also the proof of theorem 4.
Theorem 2.5 The diophantine equation 67x4 + y4 = z3 has no primitivenon-zero solutions.
Proof. From prerequisites we see that 67x, y and z are pairwise relativelyprime and x · y · z 6= 0. After congruence considerations we realize that x 6≡ymod 2 and z is odd. We have,
(y2 + x2√−67) · (y2− x2
√−67) = z3 where y2 + x2
√−67
and y2 − x2√−67 are coprime in the ring of integers (OQ
√−67) of the number
field Q√−67. Since OQ
√−67 has unique factorisation and all units (±1) are
cubes we get,
y2 + x2√−67 = (a+b
√−67
2)3 where a and b are integers of the same
parity.
Case I. a and b are even.
On some Diophantine equations ... 419
Thus y2 + x2√−67 = (c + d
√−67)3. Hence y2 = c(c2 − 201d2) (1) and
x2 = d(3c2 − 67d2) (2). c 6≡ dmod 2 since z is odd. (c, d) = 1 since (x, y) = 1and 67 - c since 67 - y.
Subcase 1.) 3 - c · dHence (c, c2−201d2) = 1 and from equation (1) we have c = ±V 2
1 (3). Since(d, 3c2−67d2) = 1 we get from equation (2) d = ±V 2
2 (4) and 3c2−67d2 = ±U21
(5) where the positive sign is rejected after examination modulo 3 of equation(5). Hence from (3), (4) and (5) we have,
3V 41 − 67V 4
2 = −U21
67V 42 − 3V 4
1 = U21 . If V2 is even and V1 is odd
we have,
67V 42 = U2
1 + 3V 41 and this is impossible since U1
is odd and U21 + 3V 4
1 6≡ 0 mod 16. If V2 is odd and V1 is even we get,
3V 41 = 67V 4
2 − U21 and this is impossible after
examination mod 16.
Subcase 2.) 3 | c and 3 - d.Hence c = 3t and from equation (1) we have,
3t(9t2 − 201d2) = y2
9t(3t2 − 67d2) = y2 and since (9t, 3t2 − 67d2) = 1we see that t = ±V 2
3 (6). Furthermore (d, 3c2 − 67d2) = 1 and from equation(2) and (6) we have since c = 3t,
d(27t2 − 67d2) = x2. Hence d = ±V 24 (7) and
27t2 − 67d2 = ±U22 (8) where the positive sign is rejected after examination
modulo 3 of equation (8). Thus from from (6), (7) and (8) we have,
27V 43 −67V 4
4 = −U22 and this equation is impossible
after examination modulo 16 since U2 is odd and V3 6≡ V4 mod 2.
Subcase 3.) 3 | d and 3 - c.Hence d = 3v and from equation (2) we have,
3v(3c2 − 67 · 9v2) = x2
9v(c2 − 201v2) = x2 and since (9v, c2 − 201v2) = 1we see that v = ±V 2
5 (9). Since (c, c2 − 201d2) = 1 and d = 3v we get fromequation (1),
420 Gustaf Soderlund
c = ±V 26 (10) and c2− 201 · 9v2 = ±U2
3 (11) where thenegative sign is rejected after examination modulo 3 of equation (11). Hencefrom (9), (10) and (11) we have,
V 46 −1809V 4
5 = U23 and according to [5] this equation
has non-zero solutions if and only if the corresponding elliptic curve Y 2 = X3−1809X has non-zero rank. This elliptic curve is not included in [6] since theconductor is too large. However after using Magma (See acknowledgements!)we find that this curve has rank zero and cosequently there are no non-zerosolutions and subcase 3 must be impossible.
Case II. a and b are odd.Hence we have,
y2 + x2√−67 = (a+b
√−67
2)3. Thus,
y2 = a(a2−201b2)8
(1) and x2 = b(3a2−67b2)8
(2) where67 - a since 67 - y and (a, b) = 1 since (x, y) = 1.
Subcase 1.) 3 - a · bHence we have (a, (a
2−201b2)8
) = 1 and (b, (3a2−67b2)8
) = 1. Thus from equation(1) we get a = ±U2
1 (3) and a2 − 201b2 = ±8V 21 (4), where the positive sign is
rejected after examination modulo 3 of equation (4). Hence,
U41 − 201b2 = −8V 2
1 (5)
Furthermore from from equation (2) we have b = ±U22 (6) and 3a2−67b2 =
±8V 22 (7), where the negative sign is rejected after examination modulo 3 of
equation (7). Hence from (3), (5), (6) and (7) we have,
U41 − 201U4
2 = −8V 21 =⇒ 201U4
2 −U41 = 8V 2
1 (8) and,
3U41 − 67U4
2 = 8V 22 (9).
From equation (8) we see that 201U2, U1 and 8V1 are pairwise relativelyprime. We now claim that equation (8) implies that 5 - U1 · U2. Assume thecontrary namely 5 | U1 or 5 | U2. If 5 | U1 we see from (8) that U4
2 = 5K1 + 1and V 2
1 = 5L1±1 so U41 = 201(5K1+1)−8(5L1±1) = 5K2+201±8 6≡ 0 mod 5
thus a contradiction. If 5 | U2 we see from (8) that U41 = 5K3 + 1 so 201U4
2 =5K3 + 1 + 8(5L1± 1) = 5K4 + 1± 8 6≡ 0 mod 5 and this is also a contradiction.If U4
1 = 5K3 +1 (10) and U42 = 5K1 +1 (11) are inserted in equation (9) we get
3(5K3+1)−67(5K1+1) = 8V 22 =⇒ 8V 2
2 = 5K5−64 =⇒ 5 - V2 =⇒ V 22 = 5L2±1
On some Diophantine equations ... 421
(12). So if the substitutions of U41 , U
42 and V 2
2 in (10), (11) and (12) are insertedin equation (9) we have,
3(5K3 + 1)− 67(5K1 + 1) = 8(5L2 ± 1). Thus,
64± 8 ≡ 0 mod 5 and this is absurd. Hence subcase1 must be impossible.
Subcase 2.) 3 | a and 3 - b.Hence a = 3t and from equation (1) we have y2 = 9t(3t2−67b2)
8(13). Since
(9t, (3t2−67b2)8
) = 1 we see from equation (13) that t = ±U23 (14) and 3t2−67b2 =
±8V 23 (15) where the negative sign is rejected after examination modulo 3 of
equation (15). Hence,
3U43 − 67b2 = 8V 2
3 (16)
Since (b, 3a2−67b28
) = 1 we see from equation (2) that b = ±U24 (17) and
3a2−67b2 = ±8V 24 (18) where the negative sign is rejected after examination
modulo 3 of equation (18). Since a = 3t we get from (17), (18) and (14),
27U43 − 67U4
4 =8V 24 (19) and from (16) and (17) we
have,
3U43 − 67U4
4 = 8V 23 (20).
From (19) and (20) we realize after congruence considerations that V4 mustbe odd and V3 must be even. (19) - (20) =⇒ 24U4
3 = 8V 24 − 8V 2
3 =⇒ V 23 =
V 24 − 3U4
3 and this is impossible since V 24 − 3U4
3 6≡ 0 mod 4.
Subcase 3.) 3 - a and 3 | b.Hence b = 3v and from equation (2) we have x2 = 9v(a2−201v2)
8(21). Since
(9v, (a2−201v2)
8) = 1 we see from (21) that v = ±U2
5 (22) and a2−201v2 = ±8V 25
(23) where the positive sign is rejected after examination modulo 3 of equation(23). Hence,
a2 − 201U45 = −8V 2
5 (24)
Since (a, (a2−201b2)
8) = 1 we see from equation (1) that a = ±U2
6 (25) anda2 − 201b2 = ±8V 2
6 (26) where the positive sign is rejected after examinationmodulo 3 of equation (26). Since b = 3v we have from (22), (25) and (26),
422 Gustaf Soderlund
U46 − 1809U4
5 = −8V 26
1809U45 − U4
6 = 8V 26 (27) and from (24) and (25)
we get,
U46 − 201U4
5 = −8V 25
201U45 −U4
6 = 8V 25 (28) and from (27) and (28) we
conclude after congruence considerations that V6 must be even and V5 must
be odd. (27) - (28) =⇒ 1608U45 = 8V 2
6 − 8V 25 . Hence,
201U45 = V 2
6 − V 25
V 26 = 201U4
5 + V 25 and this is impossible since
201U45 + V 2
5 6≡ 0 mod 4.The proof of subcase 3 is hereby completed and also the proof of theorem
5.
Theorem 2.6 If h is a prime number of the form 8m + 3 then the dio-phantine equation hx2 + y4 = z4 has no non-zero solutions.
Proof. Proved by Pocklington 1914 [14].
Lemma 2.7 If a2 − hb2 = c2, a and c odd, b even, h is an odd prime,(a, hb) = (a, c) = (hb, c) = 1 and a·b·c 6= 0 then there exist non-zero integers mand n such that a = ±(m2 +hn2), c = ±(m2−hn2) and b = 2mn. (m,hn) = 1andm 6≡ nmod 2.
Proof. See i.e. [17].
Theorem 2.8 If h ∈ {3, 11, 19, 43, 67, 163} then the diophantine equationhx4 + y4 = z5 has no primitive non-zero solutions.
Proof. We have (y2 + x2√−h) · (y2 − x2
√−h) = z5. From prerequistes
we see that hx, y and z are pairwise relatively prime, x · y · z 6= 0 and aftercongruence considerations we realize that x 6≡ ymod 2 and z is odd. Further-more y2 +x2
√−h and y2−x2
√−h are coprime in the ring of integers (OQ
√−h)
of the numberfield Q√−h. Since all units in OQ
√−h are fifth powers [17] we
have,
y2 + x2√−h = (a+b
√−h
2)5 where a and b are integers of
the same parity. With the exponent 5 on RHS it is easy to show that a andb must be even [17]. Hence,
On some Diophantine equations ... 423
y2 + x2√−h = (c + d
√−h)5 where c 6≡ dmod 2 since z is
odd. Hence,
y2+x2√−h = c5−10hc3d2+5h2cd4+(5c4d−10hc2d3+h2d5)
√−h.
Hence
y2 = c(c4− 10hc2d2 + 5h2d4) and x2 = d(5c4− 10hc2d2 +h2d4).(c, d) = 1 since (x, y) = 1 h - c since h - y.
Case I: 5 - x and 5 - y.Hence (c, c4−10hc2d2 +5h2d4) = 1 and (d, 5c4−10hc2d2 +h2d4) = 1. Thus
we have
y2 = c((±(c2 − 5hd2))2 − 5 · (2hd2)2) (1)
x2 = d((±(hd2 − 5c2))2 − 5 · (2c2)2) (2)
Hence from (1) and (2) we conclude that c = ±t2 (3), (±(c2− 5hd2))2− 5 ·(2hd2)2 = ±U2
2 (6) where the negative signs in (3) - (6)are rejected after congruence considerations regarding equations (4) and (6).Thus from (4) we have (±(c2 − 5hd2))2 − U2
1 = 5 · 4h2d4 and since U1 is oddand c 6≡ dmod 2 we see that d is even. On the other hand from (6) we have(±(hd2 − 5c2))2 − U2
2 = 5 · 4c4 and since U2 is odd c must be even and this isa contradiction since (c, d) = 1.
Case II: 5 | y and 5 - x.Hence c = 5S and from y2 = c(c4 − 10hc2d2 + 5h2d4) we have y2 =
25S((±(hd2 − 25S2))2 − 5 · (10S2)2) (1) where the two factors on RHS in (1)are coprime. Hence 25S = ±t2 (2) and (±(hd2 − 25S2))2 − 5 · (10S2)2 = ±U2
1
(3) where the negative signs in (2) and (3) are rejected after congruence con-siderations regarding equation (3). Furthermore from (2) we have S = g2 (4)and after congruence considerations we realize that S and c = 5g2 must beeven and d hereby odd.
From x2 = d(5c4−10hc2d2+h2d4) we have x2 = d((±(hd2−5c2))2−5·(2c2)2)where the two factors on RHS are coprime. Hence we get (±(hd2−5c2))2−5 ·(2c2)2 = ±U2
2 (5) and d = ±j2 (6) where the negative signs are rejected aftercongruence considerations regarding equation (5).
Subcase 1: h = 19.Since (±(hd2 − 5c2), 5 · 2c2) = 1 we see according to (5) and lemma 1 that
±(19d2 − 5c2) = ±(p2 + 5q2) (7) and 2c2 = 2pq =⇒ c2 = pq (8). N.B. if we
424 Gustaf Soderlund
had −2c2 = 2pq =⇒ c2 = (−p) · q = p · (−q) where (−p)2 = p2 and (−q)2 = q2.p 6≡ qmod 2 and (p, 5q) = 1. From (7) we have since the signs are independent,
19d2 − 5c2 = ±(p2 + 5q2) (9)
We first study equation (9) with positive sign. Hence,
19d2 − p2 = 5q2 + 5c2
If p is odd and q is even we see since d is odd and c is even that 19d2−p2 6≡0 mod 4 and 5q2 +5c2 ≡ 0 mod 4 and a contradiction is established. If p is evenand q is odd we have 19d2 − 5q2 6≡ 0 mod 4 and p2 + 5c2 ≡ 0 mod 4 and this isabsurd. Next we explore equation (9) with negative sign. Hence,
19d2 − 5c2 = −p2 − 5q2 (10)
However since c = 5S we see from (4) that c = 5g2 and from (8) we have25g4 = pq =⇒ p = B4 and q = 25A4. We first consider the case when p is oddand q is even. From equation (10) we have 19d2 +B8 = 125A4B4 − 5 · 625A8.Since d and B are odd and A is even we see that 19d2 + B8 6≡ 0 mod 16 and125A4B4− 5 · 625A8 ≡ 0 mod 16 and this is a contradiction. If p is even and qis odd we have 19d2+5 ·625A8 = 125A4B4−B8. However from (6) we see thatd = j2. Hence 19j4+5·625A8 = 125A4B4−B8 where 19j4+5·625A8 6≡ 0 mod 16and 125A4B4 − B8 ≡ 0 mod 16 and we have a contradiction again. Thus wehave shown that subcase 1 is impossible.
Subcase 2: h = 11From equation (3) we have (±(hd2 − 25S2))2 − 5 · (10S2)2 = U2
1 where((±(hd2 − 25S2), 5 · 10S2) = 1. Hence according to lemma 1 we have,
±(11d2 − 25S2) = ±(k2 + 5l2) and since the signsare independent we get,
11d2 − 25S2 = ±(k2 + 5l2) (11)
10S2 = 2kl (12) where k 6≡ lmod 2and (k, 5l) = 1. We first study equation (11) with negative sign. Hence11d2−25S2 = −k2−5l2 =⇒ 11d2 +k2 = 25S2−5l2. Thus 11d2 +k2 ≡ 0 mod 5(13). However from (12) we have 10S2 = 2kl =⇒ kl = 5S2 =⇒ k = L2.From (6) we see that d = j2. Hence 11j4 + L4 ≡ 0 mod 5. However j isodd and 5 - j and 5 - L. Thus j4 ≡ 1 mod 10 and L4 ≡ 1, 6 mod 10. Hence11j4 + L4 ≡ 2, 7 mod 10 which is a contradiction to (13).
On some Diophantine equations ... 425
Next we study equation (11) with positive sign. If k is odd and l is evenwe have 11d2 − k2 = 25S2 + 5l2. Since d and k are odd and S and l are evenwe get 11d2 − k2 6≡ 0 mod 4 and 25S2 + 5l2 ≡ 0 mod 4 and a contradictionis established. If k is even and l is odd we have 11d2 − 5l2 6≡ 0 mod 4 and25S2 + k2 ≡ 0 mod 4 and this is absurd. The proof of subcase 2 is herebycompleted.
Subcase 3: h ∈ {3, 43, 67, 163}From equation (3) we have (±(hd2−25S2))2−5·(10S2)2 = U2
1 . Furthermore((±(hd2 − 25S2)), 5 · 10S2) = 1. Hence, in compliance with lemma 1 ∃ m andn such that ±(hd2−25S2) = ±(m2 + 5n2) and since the signs are independentwe have hd2 − 25S2 = ±(m2 + 5n2) (14.), m 6≡ nmod 2 and (m, 5n) = 1.
We first examine equation (14.) with positive sign. Hence,
hd2 − 25S2 = m2 + 5n2. Hence,
hd2−m2 ≡ 0 mod 5. But d is odd and 5 - d =⇒ d2 ≡ 1, 9 mod 10.Moreover, h ≡ 3, 7 mod 10 and we must conclude that hd2 ≡ 3, 7 mod 10.5 - m =⇒ m2 ≡ 1, 4, 6, 9 mod 10. Hence,
hd2−m2 ≡ 1, 2, 3, 4, 6, 7, 8, 9 mod 10 and this contradicts thefact that hd2 −m2 ≡ 0 mod 5. Finally we study equation (14) with negativesign. Hence,
hd2 − 25S2 = −m2 − 5n2
hd2 +m2≡ 0 mod 5. Thus, according to previous discussion
we have,
hd2 + m2 ≡ 1, 2, 3, 4, 6, 7, 8, 9 mod 10 and we have again acontradiction. This completes the proof of case II.
Case III: 5 - y and 5 | xHence 5 | d =⇒ d = 5S. From x2 = d(5c4 − 10hc2d2 + h2d4) we have
where the negative signs are rejected after congruence considerations regardingthe last equation. Thus we have 25S = V 2
1 and (±(c2−25hS2))2−5·(10hS2)2 =U21 (1). Moreover S = g2 (2) and we realize after congruence considerations
that g must be even. y2 = c(c4 − 10hc2d2 + 5h2d4) where the two factors onRHS are coprime. Hence c = ±t2 (3) where t is odd. Since ((±(c2−25hS2)), 5·10hS2) = 1 we have from (1) according to lemma 1,
426 Gustaf Soderlund
10hS2 = 2mn =⇒ 5hS2 = mn and from (2) we get 5hg4 = mn (4)
±(c2 − 25hS2) = ±(m2 + 5n2) and since the signs are independentwe have,
c2 − 25hS2 = ±(m2 + 5n2) (5) where m 6≡ nmod 2 and(m, 5n) = 1. The negative sign in (5) is rejected after examination modulo 4of equation (5) since c is odd and S is even. Hence,
c2 − 25hS2 = m2 + 5n2 and from (2)and (3) we have t4 = c2 = m2 + 5n2 + 25hg4 (6). If n is odd and m is even wesee that c2 − 5n2 6≡ 0 mod 16 and from (4) we get m2 + 25hg4 ≡ 0 mod 16 andthis is a contradiction. Hence m is odd and n is even. (4) =⇒ h | m or h | n.
Subcase 1: h | mHence from (4) we have m = hA4 and n = 5B4 where A is odd and
B is even and if these values of m and n are inserted in (6) we get c2 =h2A8 + 5 · 25B8 + 25hg4. Hence c2 − h2A8 = 5 · 25B8 + 25hg4 and sincec = ±t2 according to (3) we see that t4 − h2A8 = 5 · 25B8 + 25hg4. Sinceh = 8R + 3, t and A odd, B and g even we have t4 − h2A8 6≡ 0 mod 16 and5 · 25B8 + 25hg4 ≡ 0 mod 16 and we have a contradiction.
Subcase 2: h | nHence we have from (4) m = A4 and n = 5hB4 where A is odd and B
is even. If these values of m and n are inserted in (6) we have if B = 2v,c2 = A8 + 32000h2v8 + 400hA4v4 =⇒ c2 = (A4 + 200hv4)2 − 5 · (40hv4)2 (7).But (A4 + 200hv4, 5 · 40hv4) = 1 and from (7) according to lemma 1 we have,
40hv4 = 2αβ =⇒ 20hv4 = αβ(8)
A4 + 200hv4 = ±(α2 + 5β2) andsince A4, v4, α2 and β2 are positive we get A4 + 200hv4 = α2 + 5β2 (9) whereα 6≡ βmod 2 and (α, 5β) = 1. β must be even and α must be odd. Assume thecontrary namely β is odd and α is even. Thus we have A4−5β2 = α2−200hv4.Since A and β are odd we get A4 − 5β2 6≡ 0 mod 8 and from (8) we see thatα2 − 200hv4 ≡ 0 mod 8 and a contradiction is established.
From (8) we get h | α or h | β. If h | α we have from (8) α = hL4, β = 20T 4
and LT = v. If these values of α and β are inserted in equation (9) we have,
A4 = h2L8 + 5 · (20T 4)2 − 200hL4T 4
A4 = (±(hL4 − 100T 4))2 − 8000T 8
On some Diophantine equations ... 427
A4 + 500 · (2T 2)4 = (±(hL4 − 100T 4))2. Accordingto [5] this equation has non-zero solutions if and only if the elliptic curveY 2 = X3 + 500X has non-zero rank and possible zero solutions would violatethe condition x · y · z 6= 0. In [6] we find indeed that this curve has rank zero.
Thus we have left to consider the case when h | β in equation (8). Hence,
α = L4 and β = 20hT 4. From equation (8) and (9) wehave,
A4 = L8 + 5 · 400h2T 8 − 200hL4T 4
A4 = (±(L4 − 100hT 4))2 − 8000h2T 8
A4 = (A2)2 = (±(L4 − 100hT 4))2 − 5 · (40hT 4)2 and since(±(L4 − 100hT 4), 5 · 40hT 4) = 1 we have according to lemma 1,
±(L4 − 100hT 4) = ±(ε2 + 5ϕ2) where ε 6≡ ϕmod 2 and(ε, 5ϕ) = 1. Since the signs are independent we get L4−100hT 4 = ±(ε2 +5ϕ2)and after congruence considerations we realize that the negative sign must berejected since L4 = α is odd. Hence L4−100hT 4 = ε2 +5ϕ2 (10). Furthermorealso according to lemma 1 we see that 40hT 4 = 2εϕ =⇒ εϕ = 20hT 4 (11). ϕcan not be odd and hence ε can not be even since also according to lemma 1 wehave A2 = ±(ε2−5ϕ2). Hence if ϕ is odd and ε even we must after congruenceconsiderations reject the positive sign =⇒ 5ϕ2 − A2 = ε2. However accordingto (11) 4 | ε so 5ϕ2 − A2 ≡ 0 mod 16 and this is absurd. Hence ϕ must beeven and ε must be odd. So if h | ε we have from (11) ϕ = 20V 4
2 and ε = hU46 .
Thus from (10) and (11) we have L4 − h2U86 = 100hU4
6 · V 42 + 5 · 400V 8
2 (12)where L and U6 are odd. If V2 is odd we see from (12) that RHS 6≡ 0 mod 8but LHS ≡ 0 mod 8. If V2 is even we see from (12) that RHS ≡ 0 mod 26 butLHS 6≡ 0 mod 26. Thus we have a contradiction and equation (12) must beimpossible. If h | ϕ we have from (11) ϕ = 20hV 4
2 and ε = U46 . Hence from
(10) we have,
L4 = U86 + 5 · (20hV 4
2 )2 + 100hT 4. Since 5εϕ = 100hT 4
according to (11) we get,
L4 = U86 + 5 · (20hV 4
2 )2 + 100hU46V
42
L4 = (L2)2 = (U46 + 50hV 4
2 )2 − 5 · (10hV 42 )2 and since
(U46 + 50hV 4
2 , 5 · 10hV 42 ) = 1 we have according to lemma 1,
U46 + 50hV 4
2 = ±(M2 + 5N2) and since U46 , V
42 ,M
2 and N2 arepositive we get,
428 Gustaf Soderlund
U46 + 50hV 4
2 = M2 + 5N2 (13)
2MN = 10hV 42 =⇒MN = 5hV 4
2 (14)
L2 = ±(M2 − 5N2) (15). M 6≡ N mod 2 and(M, 5N) = 1. M can not be even and N can not be odd. Indeed suppose thatM is even and N is odd. Thus in (15) we must reject the positive sign aftercongruence considerations =⇒ L2 = 5N2−M2 =⇒M2 = 5N2−L2. Howeverfrom (14) we see that 16 |M =⇒ 5N2 −L2 ≡ 0 mod 28 and this is impossible.Hence M is odd and N is even. From (14) we have h | M or h | N . If h | Mwe get M = hU4
7 and N = 5V 43 . If these values of M and N are inserted in
(13) we have since MN = 5hV 42 ,
U46 = h2U8
7 + 5 · (5V 43 )2 −50hU4
7V43
U46 − h2U8
7 = 5 · (5V 43 )2 −50hU4
7V43 . However, since V3
is even and U6 and U7 are odd we have 5 · (5V 43 )2 −50hU4
7V43 ≡ 0 mod 25 and
U46 − h2U8
7 6≡ 0 mod 25 and this is a contradiction. If h | N we have N = 5hV 43
and M = U47 . If these values of M and N are inserted in (13) we get since
MN = 5hV 42 ,
U46 = U8
7 + 5 · (5hV 43 )2 − 50hV 4
3 U47
U46 = (U2
6 )2 = (±(U47 − 25hV 4
3 ))2 − 5 · (10hV 43 )2 where
(U47 − 25hV 4
3 , 5 · 10hV 43 ) = 1. Hence according to lemma 1 we have,
±(U47 − 25hV 4
3 ) = ±(λ2 + 5µ2) and since the signs areindependent we get U4
7 − 25hV 43 = ±(λ2 + 5µ2) (16) and 2λµ = 10hV 4
3 =⇒λµ = 5hV 4
3 (17). λ 6≡ µmod 2 and (λ, 5µ) = 1 and the negative sign in (16)must be rejected after congruence considerations. Hence from (16) and (17)we have,
U47 = λ2 + 5µ2 + 25hV 4
3 (18) and λµ = 5hV 43
(19) where U7 is odd and V3 is even. However (18) and (19) are of thesame type as equation (6) and (4). Moreover U4
7 < t4 so we may proceedby infinite descent and conclude that (18) and (19) are impossible. Hencethe proof of case III subcase 2 is completed and also the proof of theorem7.
Theorem 2.9 If the diophantine equation hx4 + y4 = z3 has no primitivenon-zero solutions then so has the equation hx4 + y4 = z6.
In conclusion since h ∈ {3, 11, 19, 43, 67, 163} are prime numbers ≡ 3 mod 8and hx4 + y4 = z4 ⇐⇒ h(x2)2 + y4 = z4 theorem 2.1-2.6 and 2.8-2.9 willprove all propositions in the abstract.
Acknowledgements. The author thanks Andrej Dujella for the rankdetermination of the elliptic curve Y 2 = X3 − 1809X according to theorem 5case I subcase 3.
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Received: October 21, 2019; Published: November 15, 2019
