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I faced a doubt in this question while solving some maths problem. Please, solve it.

A natural number $n$ is chosen strictly between two consecutive perfect squares. The smaller of these two square numbers is obtained by subtracting $k$ from $n$ and the larger one is obtained by adding $l$ to $n$. Prove that $n-kl$ is a perfect square.

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Example: n is 10, k is 1 and l is 6. now $10-1*6=4=2^2$ –  LeeNeverGup Nov 29 '13 at 11:19
    
@LeeNeverGup What you are saying is true, but you can't prove a statement by giving out an example. –  CODE Nov 29 '13 at 11:37
    
@CODE Of course, the example was for helping myself & other to figure out what the statement say. –  LeeNeverGup Nov 29 '13 at 11:42
    
I'm sorry if this is not the place to ask, but may I ask you where you did encounter this problem? It was on a Belgian math contest in 2012, so I'm curious what your source is. (See vwo.be/vwo/files/finale12.pdf, question 2. It is not written in English but you can see it is the same question.) –  barto Dec 2 '13 at 22:51
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2 Answers

I will reform the question as: $a^2=n-k$ and $(a+1)^2=n+l$. So we have: $a^2-n=-k, (a^2-n)+2a+1=-k+2a+1=l$ Thus, $kl=k(2a+1-k)$ and we had: $n=a^2+k$ So, $n-kl=a^2+k-2ak-k+k^2=(k-a)^2$ which is a perfect square.

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If the numbers are $p^2$, $p^2 + k$ and $p^2 + k + l$, then we must have that $k+l = 2p+1$.

A little algebra shows that $p^2 + k - kl = (p-k)^2$, after putting $l = (2p+1) -k$.

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