There exist $a,b$ integers such that $a³+b³=311$?

Question

The title says it all. I know $a$ and $b$ can't be zero because 311 is prime. What can I do next?

Answer 1

You need to first factor $$a^3 +b^3$$= $$(a+b)* (a^2-ab+b^2)$$ so one of the factor must be 1 and another 311. From that we can say that a or b must be negative because a+b or $(a^2-ab+b^2)$ cant be 1 in other case. Can you continue now?

Answer 2

$311\equiv 3\pmod 7$

But $a^3\equiv 0,1,6\pmod 7$ for $a=0..6$ thus $a^3+b^3\in\{0,1,2,5,6\}\neq 3$

Thus it has not integer solution.

Answer 3

$$a^{3}+b^{3}=(a+b)(a^2-ab+b^2)$$ since 311 is prime number, such that there only four combinations $$a+b=1;\quad a^2-ab+b^2=311\quad\quad\quad\quad(1)$$ $$a+b=311;\quad a^2-ab+b^2=1\quad\quad\quad\quad(2)$$ $$a+b=-1;\quad a^2-ab+b^2=-311\quad\quad\quad\quad(3)$$ $$a+b=-311;\quad a^2-ab+b^2=-1\quad\quad\quad\quad(4)$$ you can solve all the four equation set, it is clear that no integers satisfy the original equation.

Answer 4

You can also do this with inequalities. Suppose WLOG, $a\geq b$. There are two cases:

Case 1: $b$ is positive. Then $a^3 \leq 311$, so $a\leq 6$. If $a=5$ then $a^3+b^3 <311$, and $a=6$ doesn't work, so there are no solutions here.

Case 2: $b$ is negative. So replace $b$ by $-b$ and write $a^3 = 311 +b^3$. Notice that $(b+1)^3 = b^3+3b^2 + 3b +1$, so if $3b^2+3b+1 >311$, there can't be a cube in the range $b^3$ to $b^3+311$.

So $b<10$. Check $311+b^3$ for $b=1, 2, \ldots, 9$ and find again, no solutions.