# The number of consecutive odd integers whose sum can be expressed as $50^2-13^2$

Here i have a question that

To find the number of consecutive odd integers whose sum can be expressed as $50^2-13^2$

Just i am unable to understand the question what is really it is asking. Please someone explain me.

• I.e., how many (2n+1)+(2n+3).. etc is equal to (63*37) – Mann Jun 4 '15 at 9:01
• I am somewhat getting it....i m trying.. – Pratyush Jun 4 '15 at 9:06
• But i think there should be the reason of giving $50^2-13^2$ – Pratyush Jun 4 '15 at 9:09

We have $$1+ 3 +...+(2n-1) = n^2$$ hence $$50^2 -13^2 =1+3 +...+99 -(1+3+...+25) =27 +29 +...+99$$
• But this is not the only solution, because $50^2-13^2$ can be expressed in other ways as a difference of two squares. For instance, $50^2-13^2 = 63 \times 37 = 21 \times 111 = 66^2-45^2$. – TonyK Jun 4 '15 at 9:45