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I know that there might be a duplicate of this. But I don't know where.

I tried equatin this to $X^2$ and and then bringing it to the other side and completing the square. What next? Is there a way to solve these types of questions?

Stefan4024
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Lenol
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  • Related: http://math.stackexchange.com/questions/524365/how-many-integer-values-of-n-are-possible-for-n225n19-to-be-a-perfect-squ?rq=1 – abiessu Nov 01 '13 at 17:14

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Your proposed procedure is right. We get $(N+10)^2-89=y^2$, that is, $x^2-y^2=89$.

So we want to solve $(x-y)(x+y)=89$. Note that the possibilities for $x-y$ are $1$, $-1$, $89$, and $-89$. Write down the corresponding value of $x+y$, and solve for $x$.

André Nicolas
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  • Cool. Is there any other ways? – Lenol Nov 01 '13 at 17:19
  • It is the standard approach for solving $x^2-y^2=a$, where $n$ is a non-zero integer. There are solutions precisely if $n$ is not of the form $4k+2$. The solutions are obtained by factoring $n$ in all possible ways as a product $ab$ of integers which are both even or both odd. For $89$, there are not many solutions since $89$ is prime. For your particular problem, one could bypass factorization by showing the solutions cannot be too large, and then doing a search. But it would take longer. – André Nicolas Nov 01 '13 at 17:26