Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$. 
Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$.

This is a homework question, and I understand that its part of Fermat's Last Theorem, but when I looked that up to try to figure out the homework, I realized that it is way further than what we've learned in class so far. We have to prove this, but the only way I can think to do it is by exhaustion, which would be extremely lengthy. Any little shove in the right direction would be appreciated. Thanks!
 A: Here’s the stupidest possible way to do it: $10^3$ is only one thousand, and there are only nine cubes less than that. They are $1$, $8$, $27$, $64$, $125$, $216$, $343$, $512$ and $729$. Since all are less than $500$ except $512$ and $729$, you only need to check whether $488$ or $271$ is on the list, and neither is.
A: Either both $x$ and $y$ are odd, or both are even. I haven't figured out the odd case yet, but here's the even case:
Even:
$(2a)^3+(2b)^3 = 10^3 \implies a^3 + b^3 = 125 \implies (a+b)(a^2 + ab + b^2) = 5 \cdot 25$. 
Note that since $x$ and $y$ are positive integers, writing the RHS as $1 \cdot 125$ wouldn't work.
Since $a+b$ is smaller than $a^2+ab+b^2$, we have $a+b = 5$. But then 
$a^2+ab+b^2 < a^2 + 2ab + b^2 = (a+b)^2 = 25$. 
So no even solution.
A: A hint of one way to prove this ...
What are the possible remainders when cubes are divided by $7$?  What is the remainder when $10^3$ is divided by $7$ (more formally what is $n$ $(0 \leq n < 7)$ such that $10^3 \equiv n \mod 7 $)? How can two of the possible remainders sum to $n$, and what does this tell us about $x$ and $y$?  
A: Just for kicks, let's show that not only are there no positive integer solutions to $x^3+y^3=10^3$, but there are no solutions if negative integers are allowed as well, i.e., no solutions with $xy\not=0$.
Let $s=x+y$ and $p=xy$.  Then 
$$10^3=x^3+y^3=(x+y)(x^2-xy+y^2)=(x+y)((x+y)^2-3xy)=s(s^2-3p)$$
so $s\mid10^3$.  Note that we must have $s\gt0$ (because $x$ and $y$ can't both be negative, and $x^2-xy+y^2$ is necessarily positive if $xy\lt0$), so there are just $16$ possibilities:  $s=2^m\cdot5^n$ with $0\le m,n\le3$.  But we have $s^3\equiv1$ mod $3$, so that cuts things in half:  we need $m+n$ to be even.  The possibilities are $s=1,4,10,250,25,100,40$, and $1000$.  But we can cut things further still.  Solving for $p$ gives
$$p={s^3-10^3\over3s}$$
In order to have $x+y=s$ and $xy=p$, we need $x$ to be an integer solution to the quadratic $x^2-sx+p=0$, which requires, at the very least, that
$$s^2-4p={4\cdot10^3-s^3\over3s}\ge0$$
This limits $s$ to be less than $10\sqrt[3]4$, which is clearly less than $25$, so only $s=1,4$, and $10$ remain.  We can rule out $s=10$ right away, since it give $p=0$ which contradicts the assumption $xy\not=0$.  For the other two cases, the discriminant of the quadratic is
$$s^2-4p={4\cdot10^3-s^3\over3s}=
\begin{cases}1333\quad\text{for }s=1\\
328\quad\text{for }s=4
\end{cases}$$
neither of which is a square.  (Note, if we were only interested in ruling out solutions with positive $x$ and $y$, the condition $p=xy\gt0$ and the formula $p={s^3-10^3\over3s}$ would have said $s\gt10$ right away.)
There may well be a quicker way to do all this; if so I'd be interested to see it.
