Find the sum of all the positive square-free integers whose divisors add up to 288.

Question

If 1^3 + 2^3 + 3^3 = m^2 where m is also an integer.
What are the next three consecutive positive integers such that the sum of their individual cubes is equal to a perfect square?

Answer 1

1158

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288=2^5*3^2
288=(1+3)*(1+71), giving the number 3*71=213
288=(1+5)*(1+47) --> 5*47=235
288=(1+11)*(1+23) --> 11*23=253
288=(1+2)*(1+3)*(1+23) -->2*3*23=138
288=(1+2)*(1+7)*(1+11) --> 2*7*11=154
288=(1+3)*(1+5)*(1+11) --> 3*5*11=165

213+235+253+138+154+165= 1158