Number theoryDifficulty 4.5AIMEProve itSilk Road Mathematics Competition
Distinct positive integers A and B are given. Prove that there exist infinitely many positive integers that can be represented both as x12+Ay12 for some positive coprime integers x1 and y1, and as x22+By22 for some positive coprime integers x2 and y2. (Golovanov A.S.)
Solution
Without loss of generality A>B. Choose an arbitrary prime p>2 and let's find x1 and x2 so that x12+A(2p)2=x22+B(2p)2, Hence, x22−x12=4Cp2 where C=A−B. Set x1=Cp2−1 and x2=Cp2+1. If x1 and x2 are both odd, then they are both coprime with y=2p, and we have x12+Ay2=x22+By2. If they are both even, then 2x1 and 2x2 are both coprime with y=p, and we have (2x1)2+Ay2=(2x2)2+By2. The number to which we have found two such forms will be not less than p2, thus proving there are infinitely many such numbers.
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