Olympiads all around!

We must solve for integers, $n^2+204\cdot n=m^2$ . Completing the square, this is $(n+102)^2=m^2+102^2.$ A Pythagorean triple! We know about these triples that

• After possibly dividing by a common factor 2, it is of the form $a^2+b^2=c^2$ with $a$ and $c$ odd and $b$ a multiple of four.

• There exist unique natural numbers $p > q$ such that $a = p^2-q^2,\ \ b=2pq,\ \ c=p^2+q^2.$

In our case, 102 is not a multiple of four so we must divide by a common factor 2 (showing that $m, n$ are even): $51 = p^2-q^2;\ \ \frac{m}{2} = 2pq;\ \ 51+\frac{n}{2} = p^2+q^2.$ Subtracting the first and last equation, we find $n = 4q^2.$ Moreover, $51 = p^2-q^2=(p+q)(p-q)$ has only two solutions, because $51=51\cdot 1=17\cdot 3$ . To maximize $n$ we need a large value of $q$ , i.e. the factors should be differ as much as possible. Thus we choose $p-q = 1, p+q = 51$ : $p = 26, q = 25;\ \ n=4q^2 = 2500;\ \ m = 4pq = 2600.$

Bonus : The only other solution is now also clear; with $p+q=17$ , $p-q=3$ we have $p = 10, q = 7;\ \ n = 4\cdot 7^2 = 196;\ \ m = 4\cdot 10\cdot 7 = 280.$

Let $\sqrt{n^{2}+204n}$ = $x$

Where $x$ is a natural number

$n^{2}+204n-x^{2}=0$

$n =\dfrac{ -204 \pm \sqrt{204^{2}+x^{2}}}{2}$

$n = -102 \pm \sqrt{102^{2}+x^{2}}$

$102^{2}+x^{2}=y^{2}$

$y^{2}-x^{2}=102^{2}$

$(y-x)(y+x)=102^{2}$

Max(x)= 2600

Therefore, $n=-102 \pm \sqrt{102^{2} +2600^{2}}$

Maximum value of n will be (obviously) when sign between the terms is $+$

$n=-102+\sqrt{102^{2}+2600^{2}}$

$n=-102+2602$

$n=\boxed{2500}$