How to prove that there is no infinite arithmetic progression of perfect squares

This question from a school Olympiad paper !

How can I prove this directly or using contradiction ?

For example : 1 ,25 , 49 are perfect squares and has 24 as the differnce ! but not infinite !

  • 1
    $\begingroup$ Hint: Consider the arithmetic progression $a,a+d,a+2d,a+3d,\dots$. If it has infinitely many perfect squares, it contains a perfect square $x^2\gt d^2$. Now look at $x^2+d$. $\endgroup$ – André Nicolas Jan 24 '16 at 16:49

When $n^2 + d = m^2$ you get $(n-m)(n+m) = d$. For any fixed $d$ this limits the possible distinct values of $n,m$ to a finite collection. From this the claim follows quite directly.


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