Problem:
Find the sum of all positive integers n such that when 13+23+33+⋯+n3 is divided by n+5, the remainder is 17.
Solution:
We are told the remainder when 13+23+⋯+n3 is divided by n+5 is 17. Recall Nicomachus’s Theorem, which states the following:
13+23+⋯+n3=(2n(n+1))2
Let S=(2n(n+1))2. We want S≡17(modn+5). Let m=n+5, so n=m−5. Substituting into the expression:
S=(2(m−5)(m−4))2
We want:
(2(m−5)(m−4))2≡17(modm)⇒m∣(2(m−5)(m−4))2−17
Try small values of m such that m>17. Check m=83:
n=78,S=(278⋅79)2=30812=9481961
Then 9481961÷83=114211 remainder 17, so this works.
Check m=166, so n=161:
S=(2161⋅162)2=130412=170070081
Then 170070081÷166=1024681 remainder 17, so this works.
Check m=332, so n=327:
S=(2327⋅328)2=536282=2876029584
Then 2876029584÷332 leaves remainder 64, so this does not work.
Thus, the valid values of n are 78 and 161, and the requested sum is:
239
The problems on this page are the property of the MAA's American Mathematics Competitions