Problem 5

Let

a

and

b

be positive integers. Show that if

4ab-1

divides

(4a^2-1)^2

, then

a=b

Step 3 of 3: Eliminate bad pairs by minimizing 2x + y

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In plain words

Whichever of $x, y$ is larger, either Step 1 shrinks $y$ to $z < x < y$ or Step 2 swaps $(x,y)$ to $(y,x)$ , strictly decreasing $2x+y$ in both cases.

(x,y) \text{ bad with } 2x+y \text{ minimal}: \quad x < y \implies 2x+z < 2x+y; \quad y < x \implies 2y+x < 2x+y

Detailed analysis

Suppose for contradiction that a bad pair exists, and choose a bad pair $(x,y)$ minimizing the positive integer $2x+y$ . If $x < y$ , Step 1 gives a bad pair $(x,z)$ with $z < x < y$ , so $2x+z < 2x+y$ , contradicting minimality. If $y < x$ , Step 2 gives the bad pair $(y,x)$ , for which $2y+x = (2x+y) - (x-y) < 2x+y$ , again contradicting minimality. Hence no bad pair exists, so $4ab-1 \mid (4a^2-1)^2$ forces $a = b$ .