Prove that $a^3+b^3=10^5$ has no positive integer solution.

Question

Prove that there do not exist two positive integers, $a$ and $b$, such that $$a^3+b^3=100\,000$$

I tried to use congruence modulo $7$ and some other modulo, but it does not seem to work. If possible, please prove it with modular congruences.

Answer 1

$$a^3+b^3=(a+b)(a^2-ab+b^2).$$ So if $a^3+b^3=10^5$, the only possible prime factors of $a^2-ab+b^2$ are $p=2$ and $p=5$.

The quadratic form $x^2-xy+y^2$ is irreducible modulo $2$ and modulo $5$ so if $2\mid(a^2-ab+b^2)$ then $2\mid a$ and $2\mid b$. Likewise if $5\mid(a^2-ab+b^2)$ then $5\mid a$ and $5\mid b$. If $$a^3+b^3=(a+b)(a^2-ab+b^2)=10^5$$ then $(a,b)=(ra',rb')$ where $r^2\mid 10^5$ and $a'^2-a'b'+b'^2=1$. As $a$ and $b$ are positive, this means $a'=b'=1$. Therefore $a=b=r$, and $a^3+b^3=2r^3\ne10^5$.

This argument rules out $a^3+b^3$ equalling any power of $10$ for $a$, $b$ integers $>0$.

Answer 2

The simplest approach is to assume $a \ge b$ and note that $a \lt 100000^{1/3} \lt 47$. Now make a spreadsheet and check them all.

Otherwise, I would factor $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a+b)^2-3ab)$ Now $a+b \ge 47$, divides $100000$, and can't be a big as $100$ so it can only be $50$ or $80$

Answer 3

The possible cubic residues modulo $19$ are:

$$1,7,8,11,12,18$$

Further, $100000\equiv 3\pmod{19}$

The only two residues whose sum is equivalent to $3$ would be $11$ and $11$.

This implies that both $a$ and $b$ are one of $5\pmod{19},16\pmod{19}$ and $17\pmod{19}$

Further suppose that $a\geq b$. This implies that $a^3\geq 50000$ and $b^3\leq 50000$

Since $(16+2\cdot 19)^3>100000$ and since $(17+19)^3<50000$, this would imply that $a^3=(5+2\cdot 19)^3=79507$ as this is the only cube of a number from one of those equivalence classes in the desired range but then $b^3=100000-79507=20493$ is not a perfect cube. Hence, no solutions exist.