Prove that $a^3+b^3=10^5$ has no positive integer solution. 
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.
 A: $$a^3+b^3=(a+b)(a^2-ab+b^2).$$
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 $>0$.
A: 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 $80$
A: 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.
