11.5. Find all natural numbers $a$ such that the product $n(n+a)$ is not a square of a natural number for any natural $n$.
Answer. $a=1,2,4$.

Solution. The fact that the product $n(n+a)$ is not a square of a natural number for $a=1,2,4$ follows from the inequalities $n^{2}<n(n+1)<n(n+2)<(n+1)^{2}$ and $n^{2}<n(n+4)<(n+2)^{2}, n(n+4) \neq(n+1)^{2}$. We will prove that for all other natural $a$, the product $n(n+a)$ is always a square of a natural number for a suitable value of $n$.

Let $a$ not be a power of two, that is, it is written in the form $a=2^{k}(2 m+1)$ for some $k \geq 0, m \geq 1$. In this case, for any natural $n$ we have $n(n+a)=n^{2}+2 \cdot 2^{k} \cdot m \cdot n+2^{k} \cdot n=\left(n+2^{k} m\right)^{2}+2^{k} \cdot n-2^{2 k} m^{2}$. Let $n=2^{k} m^{2} \geq 1$ - a natural number, then $n(n+a)=\left(n+2^{k} m\right)^{2}$ - a square of a natural number.

Consider the remaining case, when $a$ is a power of two, different from 1, 2, and 4, that is, it is written in the form $a=2^{k}$ for some $k \geq 3$. In this case, for any natural $n$ we have $n(n+a)=n^{2}+2^{k} n=n^{2}+2 \cdot 2^{k-2} \cdot n+2^{k-1} n=\left(n+2^{k-2}\right)^{2}+2^{k-1} n-2^{2 k-4}$. Let $n=2^{k-3} \geq 1$ - a natural number, then $n(n+a)=\left(n+2^{k-2}\right)^{2}$ - a square of a natural number.

Grading criteria. It is proven that $a=1,2$ are solutions to the problem: 1 point. It is proven that $a=4$ is a solution to the problem: 1 point. It is proven that for some infinite series of values of $a$ (say, for all odd numbers) the product $n(n+a)$ is a perfect square for a suitable value of $n$: 1 point. It is implied that the $a$ for which the answer remains unclear are also infinitely many.