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.

  • $\begingroup$ Example: n is 10, k is 1 and l is 6. now $10-1*6=4=2^2$ $\endgroup$ – LeeNeverGup Nov 29 '13 at 11:19
  • $\begingroup$ @LeeNeverGup What you are saying is true, but you can't prove a statement by giving out an example. $\endgroup$ – CODE Nov 29 '13 at 11:37
  • $\begingroup$ @CODE Of course, the example was for helping myself & other to figure out what the statement say. $\endgroup$ – LeeNeverGup Nov 29 '13 at 11:42
  • $\begingroup$ 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.) $\endgroup$ – punctured dusk Dec 2 '13 at 22:51

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.


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 $l = (2p+1) -k$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.