7.5. Determine the integer numbers $x$, for which the numbers $n=x^{2}+5 x+1$ and $m=x^{2}+3 x+$ 7 are simultaneously perfect squares.
## Solution:

$\operatorname{Given} x^{2}+3 x+7=k^{2} \Rightarrow 4 x^{2}+12 x+28=4 k^{2} \Leftrightarrow(2 x+3)^{2}+19=4 k^{2} \Leftrightarrow$ $(2 x+3-2 k)(2 x+3+2 k)=-19$.

a) $\left\{\begin{array}{l}2 x+3-2 k=-1 \\ 2 x+3+2 k=19\end{array} \Rightarrow 4 x+6=18 \Leftrightarrow 4 x=12 \Leftrightarrow x=3\right.$.

b) $\left\{\begin{array}{c}2 x+3-2 k=-19 \\ 2 x+3+2 k=1\end{array} \Rightarrow 4 x+6=-18 \Leftrightarrow 4 x=-24 \Leftrightarrow x=-6\right.$.

Thus $x^{2}+3 x+7$ is a perfect square for $x \in\{-6 ; 3\}$. For $x=-6 \Rightarrow x^{2}+5 x+1=$ 7 - which is not a perfect square, and for $x=3 \Rightarrow x^{2}+5 x+1=25=5^{2}$.

Answer: $x=3$.