Find integral solution to $a^3 - 1100 =b^3$ using modular arithmetic.
No integral solutions for this exist, so how to prove using modular arithmetic?
Earlier I had asked about $a$ and $b$ being raised to second power, but because of a mistake I was downvoted. This is the second part to that question.
I would probably notice that the map $x\mapsto x^3\pmod {11}$ is injective, because $0^3, 1^3, 2^3,\ldots, 10^3$ are all different $\pmod{11}$. Thus, if $a^3\equiv b^3\pmod{11}$, then $a\equiv b\pmod{11}$, in other words $11\mid a-b$.
Now where to go from there - not sure. I think a lot can be concluded by just using the arguments of "size". For example, obviously $a\gt b$, so we must have $a\ge b+11$. Imagine $b\ge 0$. then, $1100=a^3-b^3\ge (b+11)^3-b^3=33b^2+363b+1331\ge 1331\gt 1100$. This eliminates the case where $a$ and $b$ are both positive. The case where both are negative are analogous. Finally, say $a$ is positive and $b$ is negative, then we have $1100 = a^3-b^3=a^3+(-b)^3$ and so $|a|, |b|\le \sqrt[3]{1100}<11$ and this leaves just a few cases to investigate : $(a,b)\in\{(10,-1),(9,-2),\ldots,(1,-10)\}$ - out of which only half are worth checking, as the other half are the same (e.g. $10^3-(-1)^3=1^3-(-10)^3$).