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$a+n^2$ is the sum of two perfect squares for a given positive integer $a$ and all positive integers $n$. Show that $a$ is a perfect square.
At first I thought of putting a bound on the difference between perfect squares up to a point, like maybe choosing $n=c$ so that one of the squares on the right hand side is the smallest, to try and show that if $a = x^2 + k< (x+1)^2$ that there would have to be two pairs of squares with the same difference, but I can't seem to work out the details
The general solution of the equation $$x^2+y^2=z^2+w^2$$ is given by the identity with four arbitrary parameters $$(tX+sY)^2+(tY-sX)^2=(tX-sY)^2+(tY+sX)^2$$ Let $n$ be any integer so we have $$a+n^2=z^2+w^2$$Making $$n=tY-sX\\z=tX-sY\\w=tY+sX$$ we have three equations with four unknowns which in general have infinitely many solutions. Any way we have $$n^2=t^2Y^2+s^2X^2-2stXY\\z^2=t^2X^2+s^2Y^2-2stXY\\w^2=t^2Y^2+s^2X^2+2stXY$$ which implies $$z^2+w^2-n^2=t^2X^2+s^2Y^2+2stXY=(tX+sY)^2$$ Since $$a=z^2+w^2-n^2$$ we have $$a=(tX+sY)^2$$ We are done.
