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 $latex \mathbf{ \frac {11}{2} (2a + 10) = 11(a+5) }$ . a = 6 gives a perfect square.
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[…] Prove that sum of any 12 consecutive integers cannot be perfect square. Give an example where sum of 11 consecutive integers is a perfect squareSolution […]
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.