Which is the integer that, when added to either 100 or 164, results in a square number?
I. Solution: Let the unknown integer be $x$, and the two square numbers be $a^{2}$ and $b^{2}$. According to the problem statement:

$$
\begin{aligned}
& x+100=a^{2} \\
& x+164=b^{2}
\end{aligned}
$$

Subtracting (1) from (2)

$$
64=b^{2}-a^{2}=(b+a)(b-a)
$$

Both $(b+a)$ and $(b-a)$ cannot be odd, because their product (64) is even; if one is odd and the other is even, then $a$ and $b$ are not integers. Therefore, both must be even

$$
(b+a)(b-a)=\left\{\begin{array}{r}
32 \cdot 2 \\
16 \cdot 4 \\
8 \cdot 8
\end{array}\right.
$$

In the first case

$$
a=15 \quad b=17 \quad \text { and thus } \quad x=15^{2}-100=17^{2}-164=125
$$

In the second case

$$
a=6 \quad b=10, \quad \text { and thus } \quad x=6^{2}-100=10^{2}-164=-64,
$$

In the third case

$$
a=0, \quad b=8 \quad \text { and thus } \quad x=0^{2}-100=8^{2}-164=-100 .
$$

Thus, if we consider zero to be a square number, our problem has three solutions.

Beliczky Tibor (Celldömölk, Gábor Áron g. IV. o. t.)

II. Solution: (3) can also be written as:

$$
b^{2}=a^{2}+8^{2}
$$

It is known that Pythagorean triples, if the numbers are relatively prime, can be factored as follows (see Rademacher-Toeplitz: On Numbers and Figures, Tankönyvkiadó 1953, p. 88, or the second solution of problem 727, K. M. L. XIII., volume 2, p. 48):

$$
b=u^{2}+v^{2}, \quad a=u^{2}-v^{2}, \quad 8=2 u v
$$

where $u$ and $v$ are relatively prime and $u \geq v$. If the numbers of the triple are not relatively prime, then a common factor $\varrho$ can stand on the right-hand side of the above expressions, which can be any integer.

Accordingly,

$$
\varrho u v=\left\{\begin{array}{l}
1 \cdot 4 \cdot 1 \\
2 \cdot 2 \cdot 1 \\
4 \cdot 1 \cdot 1
\end{array}\right.
$$

and thus

$$
\begin{array}{ll}
a=\varrho\left(u^{2}-v^{2}\right)=1\left(4^{2}-1^{2}\right)=15, & b=1\left(4^{2}+1^{2}\right)=17 \\
a=2\left(2^{2}-1^{2}\right)=6, & b=2\left(2^{2}+1^{2}\right)=10 \\
a=4\left(1^{2}-1^{2}\right)=0, & b=4\left(1^{2}+1^{2}\right)=8
\end{array}
$$

Meskó Attila (Bp. VII., Madách g. III. o. t.)