**Prove that sum of any 12 consecutive integers cannot be perfect square. Give an example where sum of 11 consecutive integers is a perfect square**

**Discussion:** Suppose a, a+1, a+2 , … , a+ 11 are 12 consecutive integers.

Sum of these 12 integers are 6(2a + 11). This is an even integer. If it is square, it must be divisible by 4. But 6(2a+11) = 2 times odd (hence not divisible by 4). Thus it is never a perfect square.

For 11 consecutive integers the sum is \(\mathbf{ \frac {11}{2} (2a + 10) = 11(a+5) }\) . a = 6 gives a perfect square.

*Related*

all squares are in form of 4n+1 or 4n. for eg. 9 & 16. therefore given reasoning is not accurate because a whole square can be odd number

Please ‘read’ the solution given in the post.