Number Theory-- Sum of Cubes

Jan 2016
19
1
US
Let k be a prime integer such that \(\displaystyle 100 < k < 225\). How many distinct values of k exist such that \(\displaystyle k = a^{3} + b^{3}\) where a and b are both positive integers?

The answer key says 0 and I suspect this has something to do with k being prime. What's the explanation behind this?
 
Last edited:

Plato

MHF Helper
Aug 2006
22,492
8,653
Let k be a prime integer such that \(\displaystyle 100 < k < 225\). How many distinct values of k exist such that \(\displaystyle k = a^{3} + b^{3}\) where a and b are both positive integers?
The answer key says 0 and I suspect this has something to do with k being prime. What's the explanation behind this?
Can you argue that $a\ne b~?$ Realize that $a^3+b^3=(a+b)(a^2-ab+b^2)$
 
  • Like
Reactions: 1 person
Feb 2014
1,748
651
United States
You CAN attack this by brute force. The larger of a and b cannot exceed 6 so there are not many cases to test.

A more elegant way relates to the fact that the sum of two cubes can ALWAYS be factored.

$(a + b)(a^2 - ab + b^2) = a^3 - \cancel{a^2b} + ab^2 + \cancel{a^2b} - ab^2 + b^3 \implies$

$(a + b)(a^2 - ab + b^2) = a^3 + \cancel{ab^2} - \cancel{ab^2} + b^3 \implies$

$(a + b)(a^2 - ab + b^2) = a^3 + b^3.$

Now what?
 
  • Like
Reactions: 1 person
Dec 2013
2,002
757
Colombia
You can factorise $a^3+b^3=(a+b)(a^2-ab+b^2)$. So you need only prove that neither factor is equal to 1.
 
  • Like
Reactions: 1 person
Jan 2016
19
1
US
I forgot the sum of cubes could be factored. Now I realize this is the answer since prime numbers can't be factored. Thanks!
 
Dec 2013
2,002
757
Colombia
This is a slightly odd question because those limits on \(\displaystyle k\) are not necessary as far as I can see: \(\displaystyle k > 2\) is sufficient.

\(\displaystyle a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)\big((a-b)^2+ab\big)\)​

Now, since \(\displaystyle a \ge 1\) and \(\displaystyle b \ge 1\) we have \(\displaystyle a+b \ne 1\) and \(\displaystyle ab \ge 1\) so \(\displaystyle \big((a-b)^2+ab\big) \ge 1\) with equality if and only if \(\displaystyle a=b=1\). So the only way that \(\displaystyle k\) can be prime is when \(\displaystyle a=b=1\)
 
  • Like
Reactions: 2 people