MathLabs

Problem 5

Let aa and bb be positive integers. Show that if 4ab−14ab-1 divides (4a2−1)2(4a^2-1)^2, then a=ba=b.
Step 2 of 3: Prove symmetry of bad pairs via a modular identity
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In plain words

Replacing 11 by (4xy)2(4xy)^2 inside 4y2−14y^2-1 modulo 4xy−14xy-1 factors out 4y24y^2 and leaves 1−4x21-4x^2, transferring divisibility from (4x2−1)2(4x^2-1)^2 to (4y2−1)2(4y^2-1)^2.

4xy≡1(mod4xy−1)  ⟹  (4y2−1)2≡(4y2−(4xy)2)2=16y4(4x2−1)2≡0(mod4xy−1)4xy \equiv 1 \pmod{4xy-1} \implies (4y^2-1)^2 \equiv \big(4y^2-(4xy)^2\big)^2 = 16y^4(4x^2-1)^2 \equiv 0 \pmod{4xy-1}
Detailed analysis

Let (x,y)(x,y) be any bad pair. Working modulo 4xy−14xy-1, we have 1≡(4xy)21 \equiv (4xy)^2, so (4y2−1)2≡(4y2−(4xy)2)2=16y4(4x2−1)2≡0(mod4xy−1)(4y^2-1)^2 \equiv \big(4y^2 - (4xy)^2\big)^2 = 16y^4(4x^2-1)^2 \equiv 0 \pmod{4xy-1} since 4xy−1∣(4x2−1)24xy-1 \mid (4x^2-1)^2. Thus 4yx−1∣(4y2−1)24yx-1 \mid (4y^2-1)^2 with y≠xy \ne x, meaning (y,x)(y,x) is also a bad pair.