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Sum the integers, get squares? (Posted on 20070405) 

Can the sum of any 2112 consecutive positive integers be a perfect square?

Submitted by K Sengupta

Rating: 2.0000 (3 votes)


Solution:

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(p+1) + (p+2) + ... + (p+2112)
= (p+2112)(p+2113)/2  p(p+1)/2
= 2112p + 2112.2113/2 = 1056(2p + 2113).
But 1056 = 32.33 and (2p+2113) is odd, so the highest power of 2 dividing 1056(2p+2113) is 32 = 2^5.
Hence the sum of 2112 consecutive positive integers cannot be a perfect square.

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