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

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Solution

Let $u$ be a natural number such that $u^{2} < n < (u + 1)^{2}$ . Then $n - k = u^{2}$ and $n + l = (u + 1)^{2}$ . Thus
$n k l = n (n u^{2}) ((u + 1)^{2} n)$
$= n n (u + 1)^{2} + n^{2} + u^{2} (u + 1)^{2} n u^{2}$
$= n^{2} + n (1 (u + 1)^{2}) u^{2}) + u^{2} (u + 1)^{2}$
$= n^{2} + n (12 u^{2} 2 u 1) + u^{2} (u + 1)^{2}$
$= n^{2} 2 n u (u + 1) + (u (u + 1))^{2}$
$= (n - u (u + 1)^{2}$ .

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

Let u be a natural number such that u2<n<(u+1)2 . Then n−k=u2 and n+l=(u+1)2 . Thusnkl=n(nu2)((u+1)2n) =nn(u+1)2+n2+u2(u+1)2nu2 =n2+n(1(u+1)2)u2)+u2(u+1)2 =n2+n(12u22u1)+u2(u+1)2 =n22nu(u+1)+(u(u+1))2 =(n−u(u+1)2 .

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