<s>[INST] 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.
 [/INST] $$a^3+b^3=(a+b)(a^2-ab+b^2).$$
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 $>0$.
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