Consecutive Integers whose Sums of Squares of Divisors are Equal

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Theorem

The only two consecutive positive integers whose sums of the squares of their divisors are equal are $6$ and $7$.


Proof

The divisors of $6$ are

$1, 2, 3, 6$

and so the sum of the squares of the divisors of $6$ is:

$1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$


The divisors of $7$ are

$1, 7$

and so the sum of the squares of the divisors of $7$ is:

$1^2 + 7^2 = 1 + 49 = 50$


It remains to be shown that there are no more:



Sources