Solution September 11, 2007


Assume for the sake of contradiction that we have a and b such that a \neq b and

4ab-1 | (4a^2-1)^2.\,


4ab-1 | b^2(4a^2-1)^2 - (4ab-1)n\,

With n = 4a^3b + a^2 - 2ab\, this gives

4ab-1 | (a-b)^2\,


k = \frac{(a-b)^2}{4ab-1}. \qquad\qquad\mbox{(1)}

k\, can not be 0\, so we have k>0\,. Choose a\, and b\, among the solutions of (1) such that a\geq b\, and a\, is minimal.

Then, on the one hand,

k(4ab - 1) = a^2 - 2ab + b^2\,
-k = (a - tb)(a - \frac 1t b)

where t = 2k+1 + 2\sqrt{k(k+1)}. Since a\geq b, the second term is positive, and

a = \frac{-k}{a - \frac 1t b} \ge \frac{-k}{(1-\frac1t)b} + tb  \ge -2k +  2\sqrt{k(k+1)}b + (2k+1)b > (2k+1)b.

On the other hand, if a\, is a solution of (1) then so is a' = 2(2k+1)b - a\,. By the minimality condition, we have a\leq a'\,, i.e.

a \leq (2k+1)b\,

This is a contradiction to the previous inequality, completing the proof.