# If there is one perfect square in an arithmetic progression, then there are infinitely many

Consider the following positive integers: $$a,a+d,a+2d,\dots$$ Suppose there is a perfect square in the above list of numbers. Then prove that there are infinitely many perfect square in the above list. How can I do this?

At first I started in this way: Let the $n$th term is perfect square. Therefore, $$t_n=a+(n-1)d=m^2.$$ Then I think that I will put values at the position of $n$. But I failed to find anything from this level. Can somebody help me?

• If $m^2$ is in the sequence, so too are $(m+d)^2, (m+2d)^2, (m+3d)^2, \ldots$ – Henry Jan 9 '17 at 17:09

## 4 Answers

Note that $$(m+d)^2=m^2+(2m+d)\cdot d$$

Let $n^2$ be the known square.

Then $n^2+kd=m^2$ is equivalent to $kd=(m-n)(m+n)$. You can take $m-n$ to be any multiple of $d$, and $k$ follows.

Starting from where you left, suppose the $k$th term is a perfect square such that $$a_k = a_1 + (k-1)d = p^2$$

Now add $2pmd + m^2d^2$ to both sides where $m$ is a natural number giving us, $$\Rightarrow a_1 + (k-1)d + 2pmd + m^2d^2 = p^2 + 2pmd + m^2d^2$$ $$\Rightarrow a_1 + [(k-1)+2pm+m^2]d = (p+md)^2$$

The RHS is a perfect square, and the left side is the $(k-1+2pm+m^2)$th term for infinitely many values of $m$. Hope it helps.

• last term inside square brackets should be $m^2d$? – hypergeometric Jan 9 '17 at 17:29

Suppose that given $a_n = a+n*d = x_n^2$ there is theoretically another value $a_n' = a+n'*d = x_n'^2$ where all numbers are whole.

Taking the difference between these number yields: $$a_n' - a_n = a_n'- a_n\\ (a+n'*d) - (a+n*d) = x_n'^2 - x_n^2\\ (n' - n)d = (x_n' - x_n)(x_n' + x_n)$$ Setting the outer expressions equal to each other, and the inner expressions equal to each other yields: $$x_n' - x_n = d \\x_n' = d + x_n$$ and $$n' - n = x_n'+x_n\\n' = n + x_n'+x_n\\n' = n + d + 2 x_n$$

Therefore: If $a_n = a + n*d = x_n^2$ is a square, you can use theses values to generate a new larger square within the sequence where: $a_n' = a + n'd = a + (n+d+2*x_n)*d$. This will be: $x_n'^2 = (d+x_n)^2$.

• Please use LaTeX formatting to improve readability! – The Count Jan 9 '17 at 20:48