If 2n2 + 1 or 2n2 - 1 is a perfect square say m2 = 2n2 + 1 or 2n2 - 1 then prove that (mn)2 is a triangular number.


I tried solving for m2 or n2 and tried to bring it to the required form for a triangular number, but alas was unable to do so. However, i also realised that m is odd and tried to substitute m as 2l + 1 for some l and tried solving which didn't help as well. I have verified this for m = 3 and n=2 which in turn gives mn = 6 and in turn (mn)2 = 36which indeed is a triangular number.


I have no idea about Pell's equation as I have seen several problems on triangular numbers being solved by Pell's equation. All the topics in which I can consider myself to have sufficient knowledge include Modular arithmetic, properties of triangular number and other basic stuff such as induction,etc..

Thanks in advance!!!


$(mn)^2=m^2n^2=(2n^2+1)n^2=\frac{(2n^2+1)2n^2}{2}$ which is a triangular number. Similarly you can deal with the case $m^2=2n^2-1$

  • $\begingroup$ ohh thanks that was a veryp easy one,silly to have messed u $\endgroup$ – saisanjeev Feb 8 '18 at 12:20
  • $\begingroup$ messed up*, sorry fr d typo $\endgroup$ – saisanjeev Feb 8 '18 at 12:20

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