Solution of One of the Toughest question of Mathematics Olympiad ever

SOLUTION OF Toughest Question of Maths ever

Let us denote the square of an integer by k2$k^2$. We have to show that

a2+b2ab+1=k2$$\frac{a^2+b^2}{ab+1}=k^2$$

for some a>0$a>0$ and b>0$b>0$ where a,b,>k∈Z$a,b,>k\in \mathbb{Z}$.

Rewriting the equation:

a2+b2=k2(ab+1)=k2ab+k2$$a^2+b^2=k^2(ab+1)=k^{2}ab+k^2$$

Let k=b$k=b$:

a2+b2=b3a+b2⇒a2=b3a⇒a=b3$$a^2+b^2=b^{3}a+b^2\Rightarrow a^2=b^{3}a\Rightarrow a=b^3$$

To check, let's substitute these relations back in the original expression:

a2+b2ab+1=(k3)2+(k)2(k3)(k)+1=k6+k2k4+1=k2(k4+1)k4+1=k2