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 1 of 3: Define bad pairs and prove the shrinking property when x < y
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In plain words

Modulo 4x4x, both 4xy−14xy-1 and 4x2−14x^2-1 are −1-1, so the quotient rr is also −1(mod4x)-1 \pmod{4x} and smaller than 4x2−14x^2-1 whenever x<yx < y.

r=(4x2−1)24xy−1≡−1(mod4x)  ⟹  r=4xz−1 with 1≤z<xr = \frac{(4x^2-1)^2}{4xy-1} \equiv -1 \pmod{4x} \implies r = 4xz-1 \text{ with } 1 \le z < x
Detailed analysis

Call a pair (x,y)(x,y) of positive integers bad if 4xy−1∣(4x2−1)24xy-1 \mid (4x^2-1)^2 and x≠yx \ne y. Suppose (x,y)(x,y) is bad with x<yx < y, and let r=(4x2−1)24xy−1r = \dfrac{(4x^2-1)^2}{4xy-1}. Since r(4xy−1)=(4x2−1)2≡1(mod4x)r(4xy-1) = (4x^2-1)^2 \equiv 1 \pmod{4x} and 4xy−1≡−1(mod4x)4xy-1 \equiv -1 \pmod{4x}, we have r≡−1(mod4x)r \equiv -1 \pmod{4x}, so r=4xz−1r = 4xz-1 for some positive integer zz. Because x<yx < y, 4xz−1=(4x2−1)24xy−1<(4x2−1)24x2−1=4x2−14xz-1 = \dfrac{(4x^2-1)^2}{4xy-1} < \dfrac{(4x^2-1)^2}{4x^2-1} = 4x^2-1, which gives z<xz < x; since 4xz−1=r4xz-1 = r divides (4x2−1)2(4x^2-1)^2 and z<xz < x, the pair (x,z)(x,z) is also bad.