Mongolian Mathematical Olympiad

Suppose that arithmetic progression $a_n$ satisfies given condition and let $d$ be common difference of the progression.

If $d=0$ then $a_n=0,\forall n\in \mathbb{N}$ or $a_n=1,\forall n\in \mathbb{N}$.

Let $d\ne 0$. $a_1^2,a_2^2,a_3^2$ are terms of the given progression $\iff \exists m,n,k\in \mathbb{N},a_1^2=a_1+md,a_2^2=a_1+nd=(a_1+d{)}^2,a_3^2=a_1+kd=(a_1+2d{)}^2\iff$ $$\{\begin{array}{l}2a_1+d=n-m\\ 4a_1+4d=k-m\end{array}\Rightarrow$$ $$a_1=\frac{4n-3m-k}{4},d=\frac{m-2n+k}{2}\in \mathbb{Q}.$$ On the other hand, $a_1^2=a_1+md\iff a_1^2+(2m-1)a_1-m(d+2a_1)=0$ from where follows $a_1$ is rational root of the polynomial 2th degree with integer coefficients $P(x)=x^2+(2m-1)x-m(n-m)$. If the equation $P(x)=0$ has rational solution then the solution is integer too (coefficient of the leading term of the polynomial $P(x)$ equals to 1). Consequently $a_1\in \mathbb{Z}$. Thus $d=n-m-2a_1\in \mathbb{Z}$ and this completes the proof.