Wednesday, October 23, 2019

Sum of squares of 5 consecutive natural numbers!

Here's a difficult problem of proving that a given expression is not a perfect square. A direct approach will be a nightmare ( even not sure it can be proved) but use of a simple property of perfect squares will ease the problem. We know that any perfect squares leaves a remainder either 1 or 0 when being divided by 3 or 4. This simple result would be used to solve the problem.

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