\section*{Problem 1 - 251031}

Determine all ordered pairs \((a ; b)\) of integers \(a\) and \(b\) that satisfy the equation \(a+b=(a-b)^{2}\)!
}

Assume \((a, v)\) is a pair of integers that satisfy the given equation. Since \(a\) and \(b\) are integers, their difference is also an integer \(g\).

Thus, \(a - b = g\) and \(a + b = g^2\), and therefore

\[
a = \frac{1}{2} \left(g^2 + g\right) = \frac{1}{2} g(g + 1) \quad ; \quad b = \frac{1}{2} \left(g^2 - g\right) = \frac{1}{2} g(g - 1)
\]

Hence, the pairs

\[
\left(\frac{1}{2} g(g + 1) ; \frac{1}{2} g(g - 1)\right)
\]

with integer \(m\) satisfy the equation given in the problem. They indeed satisfy this equation, as

\[
\frac{1}{2} g(g + 1) + \frac{1}{2} g(g - 1) = g^2 = \left(\frac{1}{2} g(g + 1 - g + 1)\right)^2
\]

and both \(\frac{1}{2} g(g + 1)\) and \(\frac{1}{2} g(g - 1)\) are integers.