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Let $D_n$ be a discriminant of quadratic equation $x^2+p_{n}x+q_n=0$, for each $n\in \mathbb{N}$, i.e. $$D_n=p_n^2-4q_n.$$ By assumption we conclude that $D_1$ is a square of an integer. Further, we have: $$D_{n+1}=p_{n+1}^2-4q_{n+1}=(p_n+1{)}^2-4\left(q_n+\frac{1}{2}{p}_n\right)=p_n^2-4q_n+1=D_n+1.$$ Let's suppose that the equation $x^2+p_{n}x+q_n=0$ has two integer solutions for some $n$. Then $D_n=k^2$ for some integer $k$, hence, we have: $$D_{n+2k+1}=D_n+2k+1=k^2+2k+1=(k+1{)}^2.$$ Moreover, since $\frac{1}{2}(-p_n+\sqrt{D_n})$ and $\frac{1}{2}(-p_n-\sqrt{D_n})$ are integers, we conclude that $p_n$ and $D_n$ are of the same parity. But, $$p_{n+2k+1}\equiv p_n+2k+1\equiv p_n+1\equiv D_n+1\equiv D_n+2k+1\equiv D_{n+2k+1}(\mathrm{mod}2)$$ so the equation $x^2+p_{n+2k+1}x+q_{n+2k+1}=0$ also has two integer solutions. Hence, we proved that there are infinitely many integers $n$ for which the equation $x^2+p_{n}x+q_n=0$ has two integer solutions.