\section*{Problem 4 - 071034}

We are looking for all triples of natural numbers \(a_{i}(i=1,2,3)\) that satisfy the equation

\[
\sqrt{2 a_{1}^{2}-2 a_{2}^{2}}=a_{3}
\]

and for which \(1 \leq a_{i} \leq 10\) also holds!
}

The equation is evidently equivalent to \(2\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)=a_{3}^{2}\).

Since 2 is a prime number, it follows from \(2 \mid a_{3}^{2}\) that \(2 \mid a_{3}\), so the right-hand side of the equation is divisible by 4. Therefore, \(\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)\) must also be even, and thus at least one of these two factors must be even. Since both factors can only be even (or neither can be even) simultaneously, the right-hand side of the equation is actually divisible by 8, so due to the uniqueness of prime factorization, \(a_{3}\) must be divisible by 4.

Thus, two cases remain: \(a_{3}=4\) and \(a_{3}=8\).

1. Case: \(a_{3}=4\).

Then \(\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)=8\), and since both factors are even, due to \(a_{1}+a_{2}>0\) also \(a_{1}-a_{2}>0\) and finally \(a_{1}+a_{2}>a_{1}-a_{2}\), so \(a_{1}+a_{2}=4\) and \(a_{1}-a_{2}=2\). The solution triplet \(\left(a_{1}, a_{2}, a_{3}\right)=(3,1,4)\) is obtained, which is confirmed by substituting into the original equation.

2. Case: \(a_{3}=8\).

Then \(\left(a_{1}-a_{2}\right)\left(a_{1}+a_{2}\right)=32\). The following possible decompositions into two positive and even factors arise:

*) \(a_{1}+a_{2}=16\) and \(a_{1}-a_{2}=2\), leading to \(a_{1}=9\) and \(a_{2}=7\).

*) \(a_{1}+a_{2}=8\) and \(a_{1}-a_{2}=4\), leading to \(a_{1}=6\) and \(a_{2}=2\).

Both possible solutions are confirmed by verification.

Summary: In the considered range, there are exactly three solution triplets, namely \((3,1,4),(6,2,8)\) and \((9,7,8)\).