<s>[INST] Is there a way to show that $x^2=1$ is the only solution of $x^2+3=k^2$ for $x,k\in\mathbb{N}$ The question i'm trying to solve is: The sum of all natural numbers a such that $a^2-16a+67$ is a perfect square is (a)10, (b)12, (c)16, (4)22.
$a^2-16a+67$ 
= $a^2-16a+64+3$ 
=$(a-8)^2+3$
This is a perfect square when $(a-8)^2=1$ and so $a=7,9$ 
So the answer is $7+9 = 16$, which agrees with the answer given . 
But i don't know how to show that these are the only possible values of a, or rather that $x^2=1$ is the only solution for this $x^2+3=k^2$ is the only value that works, can i get any tips/directions? Thanks to everyone in advance.
 [/INST] If $k,x \in \Bbb N$ and $3=k^2-x^2$ 
then $k^2>x^2\,$ so $k>x\,$ so $k\ge x+1\,$ so $k^2\ge (x+1)^2$ 
so $3=k^2-x^2\ge (x+1)^2-x^2=2x+1$ 
so $3\ge 2x+1.$
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