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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$\begingroup$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.
$\endgroup$ $\begingroup$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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