It is well known that, for each positive integer nnn:
13+23+...+n3=n2(n+1)241^3 + 2^3 + ... + n^3 = \frac{n^2(n + 1)^2}{4}13+23+...+n3=4n2(n+1)2
and so is a square. Determine whether or not there is a positive integer mmm such that
(m+1)3+(m+2)3+...+(2m)3(m + 1)^3 + (m + 2)^3 + ... + (2m)^3(m+1)3+(m+2)3+...+(2m)3
is a square.
[UKMT BMO 201820182018 Round 222, Q333]
Note by Yajat Shamji 7 months ago
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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\sin \theta
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You have until 1st December, 3:00pm!