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Solution to puzzle 74: Sum of 9999 consecutive squares

Show that the sum of 9999 consecutive squares cannot be a perfect power.
That is, show that (n + 1)2 + ... + (n + 9999)2 = mr has no solution in integers n, m, r > 1.


The sum of any 9 consecutive squares is congruent, modulo 9, to
02 + 12 + ... + 82 congruent to 2(12 + 22 + 32 + 42) congruent to 6.  (Since k2 congruent to (9 − k)2 (mod 9).)
Hence the sum of 9999 = 1111 · 9 consecutive squares is congruent, mod 9, to 1111 · 6 congruent to 6.
That is, mr congruent to 6 (mod 9).
So mr is divisible by 3, but not by 32 = 9, and hence cannot be a perfect power.

Therefore, the sum of 9999 consecutive squares cannot be a perfect power.

Source: (Adobe) Portable Document Format Problems in Elementary Number Theory (problems since taken down); based on problem G 24

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