# Number Theory-- Sum of Cubes

#### cake27

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?

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#### Plato

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

1 person

#### JeffM

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?

1 person

#### Archie

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.

1 person

#### cake27

I forgot the sum of cubes could be factored. Now I realize this is the answer since prime numbers can't be factored. Thanks!

#### Archie

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 $$\displaystyle a=b=1$$

2 people