



 
We have two responses for you Randy, Try this experiment: Pick an integer n, for example say n = 13, and extend n by an extra digit by putting a 5 after it (to get 135 in my example). Now compare the square of that number with n(n+1). (In my example you would compare 13*14 with 135^{2}.) Chris
Randy, So you want n(n+1)x100 + T to be a square, where T is a two digit number. Expanding, you get 100n^{2} + 100n + T, that is (10n)^{2} + 2 × (10n) × 5 + T, which you want to be a square (A+B)^{2} = A^{2} + 2 × A × B + B^{2}. Claude  


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