We have to prove that the number

$$N=512^3 + 675^3 + 720^3$$

is composite.

I tried to use the identity $(a^3+b^3+c^3)=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc$ hoping to take out some common factors from the R.H.S. but it didn't work. I also used $a^3+b^3=(a+b)(a^2-ab+b^2)$ (in all possible combos) and tried to combine with $c^3$ but that too didn't work. I nearly spent about 5 hours struggling with the question but no result :(
Please help!
Thanks in advance.

  • 6
    $\begingroup$ cms.math.ca/Concours/MOCP/98-99/ps-4.html $\endgroup$ Jun 11, 2014 at 18:23
  • $\begingroup$ Should be posted as the answer. $\endgroup$ Jun 11, 2014 at 18:26
  • 3
    $\begingroup$ Even sweeter : sciencechatforum.com/viewtopic.php?f=19&t=5975 $\endgroup$ Jun 11, 2014 at 18:27
  • $\begingroup$ Kudos to the previous commenters for not going to the smart-aleck answer. You left the door open to it by neglecting to mention the stricture that you can't use a calculator or computer to factorize it. In the mathlete context, one would waste too much time looking for the least prime factor, which in this case is the 50th prime. $\endgroup$ Jun 12, 2014 at 2:43

1 Answer 1


Well there may be so many ways to answer. One of my teachers in a math-camp gave the following way out. I think its a good one.

Let $a=512, b=675, c=720$. Now the number looks like $a^3+b^3+c^3$ but we require a sort of $3abc$ term to resolve it into factors. First we factorize $a,b,c$ into prime factors. So, $a=2^9, b=3^3\times 5^2, c=2^4\times 3^2\times 5$. Now it can be seen that $2c^2=3ab$. Hence, $a^3+b^3+c^3=a^3+b^3-c^3+2c^2c$ and thus the problem can be solved. It can then be seen that the given number is composite.

I have posted the same question in Q & A style here. I did not see your question earlier. One of the users gave the link to your question in a comment on my question.


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