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!
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.
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 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 $y$?
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.
