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Suppose $2013+ a^2 = b^2$, $a$ and $b$ being natural numbers, what is the minimum possible value of their product, $ab$?

I have tried algebraic manipulations such as moving $a^2$ to the other side, i.e., $2013 =b^2 - a^2$.

Now I need to find two such numbers such that their product is minimum but the square of one is surely larger than $2013$... I could not find any such solution. How can I do it?

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Note that $3\cdot 11\cdot 61=b^2-a^2=(b-a)(b+a)$ with positive integers $a$ and $b$. This leaves not too many possibilities for $a$ and $b$.

Edit: If you still have found no solution, try $2013+14^2=2209=47^2$.

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Write $b=a+k$. Then the equation becomes $$2013+a^2=a^2+2ak+k^2 \Rightarrow 2013 =k(2a+k)$$

Thus $k$ is a divisor of 2013.

This is actually the same solution as Dietrich's, just written differently.

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    $\begingroup$ yes thank you very much. $\endgroup$ – sugatasen Oct 13 '14 at 16:04
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    $\begingroup$ @sugatasen You are welcome. BTW, the easiest solution to find for $a^2-b^2=$ odd integer is when $a,b$ are consecutive integers. In that case $(n+1)^2-n^2=2n+1$, and the equation $2n+1=2013$ is very easy to solve.... $\endgroup$ – N. S. Oct 13 '14 at 16:34
  • $\begingroup$ Wait, I don't seem to quite understand this proposition...where is it coming from ...can you please refer some source(book) useful for preparing for such problems? also, what book might i find this type of problem and such propositions in?? Do help out. $\endgroup$ – sugatasen Oct 14 '14 at 15:33

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