Problem:
Let a,b,c,d,e be distinct positive integers such that a4+b4=c4+d4=e5. Show that ac+bd is a composite number.
Solution:
We approach indirectly by assuming that p=ac+bd is a prime. By symmetry, we may assume that max{a,b,c,d}=a, then because a4+b4=c4+d4, we infer that min{a,b,c,d}=b. Note that ac≡−bd(modp), implying that a4c4≡b4d4(modp). Consequently, we have
b4d4+b4c4≡a4c4+b4c4=c8+c4d4(modp)
from which it follows that (c4+d4)(b4−c4)≡0(modp). Thus, p divides at least one of b−c,b+c,b2+c2,c4+d4. Because p=ac+bd>c2+b2, and −(b2+c2)<b−c<0 (because b and c are distinct), p must divide c4+d4=e5. Thus p5=(ac+bd)5 divides c4+d4, which is clearly impossible because it is evident that (ac+bd)5>c4+d4.
OR
A stronger result is possible:
Claim. Suppose a,b, and e are positive integers such that a4+b4=e5. Then a and b have a common prime factor.
Proof. Suppose on the contrary that gcd(a,b)=1. If e is even, then this forces a and b to both be odd, so a4+b4≡2(mod8) and e5≡0(mod8), a contradiction. Thus e is odd. Note for use below that 5 cannot divide both a and b, so we may assume without loss that 5 does not divide a (swapping the roles of a and b if necessary).
Factoring over the Gaussian integers we find a4+b4=(a2+ib2)(a2−ib2) and gcd(a2+ ib2,a2−ib2)=gcd(a2+ib2,2a2). But gcd(a,b)=1 implies no prime factor of a can divide a2+ib2 and e odd implies no prime factor of 2 divides a2+ib2. Thus these factors are relatively prime, and hence both are a unit multiplied by a fifth power. Since every unit in the Gaussian integers is a fifth power, that means both factors are fifth powers, or
a2+ib2=(r+is)5=r5+5ir4s−10r3s2−10ir2s3+5rs4+is5
Thus
a2b2=r(r4−10r2s2+5s4), and =s(s4−10r2s2+5r4)
Note that since gcd(a,b)=1,gcd(r,s)=1. Also since 5 does not divide a, it also does not divide r. Since
gcd(r,r4−10r2s2+5s4)=gcd(r,5s4)=gcd(r,5)=1
r must be a perfect square and we have found an integer solution (x,y,z)=(r,a/r,s) to
y2=x4−10x2z2+5z4
with gcd(x,z)=1. The following Lemma will then complete the proof of the claim.
Lemma. There are no nontrivial (z=0) integer solutions to
y2=x4−10x2z2+5z4
Proof. Suppose (x,y,z) is a solution in the positive integers with minimal z. Note that this implies that x,y,z are pairwise relatively prime. (The only case that takes a little work is that if a prime p divides x and y, then p2 divides 5z4, hence p also divides z. But then p4 divides x2 so p2 divides x and (x/p2,y/p,z/p) is a smaller solution.) Rewrite this as
20z4=(x4−5z2)2−y2=(x2−5z2+y)(x2−5z2−y)
The two factors on the right have the same parity and their product is even. Hence both are even. Any common factor p of 2x2−5z2+y and 2x2−5z2−y would have p2∣5z4, hence p∣z, and p∣∣∣∣2x2−5z2+y−2x2−5z2−y=y, a contradiction. Thus these factors of 5z4 are relatively prime. Hence they must be ±v4 and ±5w4 for some relatively prime v and w with vw=z. Then
x2−5v2w2=x2−5z2=2x2−5z2+y+2x2−5z2−y=±v4±5w4
or
x2=±v4+5v2w2±5w4
If v and w both odd, then the right hand side is either 1+5+5≡3(mod8) or −1+5−5≡7 (mod8), neither of which is possible for a square like the left hand side. Hence one of v and w is even, and in either case we get x2≡±1(mod4). Thus we must have the plus sign and
x2=v4+5v2w2+5w4
This is not the equation we started with, so we repeat the argument above (with a few changes). Rewrite this new equation as
5w4=(2v2+5w2)2−4x2=(2v2+5w2+2x)(2v2+5w2−2x)
There are two very similar cases depending on whether w is odd or even. These cases can be forced together, but we prefer to be more clear and keep them separate.
If w is odd, then the two factors on the right are both odd and any common (odd) prime factor p would have p2∣5w4, hence p∣w, and p∣(2v2+5w2+2x)−(2v2+5w2−2x)=4x, hence p∣x. But then p also divides v and we get a contradiction. Thus these factors of 5w4 are relatively prime and so must be ±t4 and ±5u4 for some relatively prime t and u with tu=w. Then
4v2+10t2u2=4v2+10w2=(2v2+5w2+2x)+(2v2+5w2−2x)=±(t4+5u4)
The left hand side is positive, so we must have the plus sign, and hence
(2v)2=t4−10t2u2+5u4
Thus (t,2v,u) is a solution to the original equation. Since u∣w and w∣z, we either have u<z (contradicting the minimality of z ) or u=z and hence t=v=1 (giving nonsense 4=1−10u2+5u4≡1(mod5)). Thus this case cannot occur.
If w is even, then the two factors are even, congruent mod 4 , and their product is divisible by 16. Hence both are multiples of 4 . Any common prime factor p of 42v2+5w2+2x and 42v2+5w2−2x would have p2∣5(w/2)4, hence p∣w, and p∣∣∣∣42v2+5w2+2x−42v2+5w2−2x=x. But this would mean p∣v, a contradiction. Thus 42v2+5w2+2x and 42v2+5w2−2x must be ±t4 and ±5u4 for some relatively prime t and u with 2tu=w. Then
v2+10t2u2=v2+25w2=42v2+5w2+2x+42v2+5w2−2x=±(t4+5u4)
Again, the left hand side is positive, so we must have the plus sign, and hence
v2=t4−10t2u2+5u4
Thus (t,v,u) is a solution to the original equation and since 2u∣w and w∣z, we have u<z. This contradicts the minimality of z and completes the proof of the lemma.
Remark. One can use essentially the same argument to show that any nontrivial integer solution to x2+y4=z5 has gcd(x,y)>1. In this case one cannot assume 5 does not divide r so there is a second case where r=5q2. Then (x,y,z)=(s,a/(5q),q2) is a solution to
y2=x4−50x2z2+125z4
This Diophantine equation also has no nontrivial integer solutions and the proof is nearly identical to the proof of the Lemma above. This stronger result was (apparently) first proven by Nils Bruin (1999). This result is at least tangentially related to Beal's conjecture.
The more general solution is due to Richard Stong.
The problems on this page are the property of the MAA's American Mathematics Competitions