Question

Let ${\left(a_n\right)}_{n\ge 1}$ be an arithmetic sequence such that $a_1^2,a_2^2,$ and $a_3^2$ are also terms of the sequence. Prove that the terms of this sequence are all integers.

Answer 1

Let d be the common difference. If d=0, then it is not difficult to see that either $a_n=0$ for all n,or $a_n=1$ for all n.

Suppose that $d\ne 0,$ and consider the positive integers m, n, p, such that $a_1^2=a_1+md,(a_1+d^2)=a_1+nd,$ and ${(a_1+2d)}^2=a_1+pd.$

Subtracting the first equation from the other two yields $\{\begin{array}{cc}2a_1+d& =n-m,\\ 4a_1+4d& =p-m\end{array}$ and solving for $a_1$ and d we obtain $a_1=\frac{1}{4}(4n-3m-p),d=\frac{1}{2}(m-2n+p),$

hence both $a_1$ and d are rational numbers.

Observe that the equation $a_1^2=a_1+md$ can be written as $a_1^2+(2m-1)a_1-m(n-m)=0,$ or,

$a_1^2+(2m-1)a_1-m(n-m)=0,$

hence $a_1$ is the root of the polynomial with integer coefficients $P\left(x\right)=x^2+(2m-1)x-m(n-m).$

By the integer root theorem it follows that $a_1$ is an integer, hence $d=nm2a_1$ is an integer as well. We conclude that all terms of the sequence are integer numbers.