Proof by cases to show no solutions in positive integers to the equation $x^4+y^4 = 100$ Use proof by cases to show that there are no solutions in positive integers to the equation $x^4+y^4 = 100$
I am looking for the logic on how to go about this problem, as i am trying to understand how to do it on my own. 
What should i prove first? 
 A: We see that $$x^4\leq100,$$
which gives
$$1\leq x\leq\sqrt{10}$$ or
$$1\leq x\leq 3.$$
Now, easy to see that for $x\in\{1,2,3\}$ we can not get an integer $y$ for which
$$x^4+y^4=100.$$
For $x=1$ we get $y^4=99$,
for $x=2$ we get $y^4=84$ and
for $x=3$ we get $y^4=19$.
Done!
A: Without loss of generality, we can assume $x\le y$. Then:
$$2x^4\le x^4+y^4=100\le 2y^4 \Rightarrow x\le2.7 \ and \ y\ge 2.7.$$
Case 1: $x=1$, then $y^4=100-1^4=99 \Rightarrow \emptyset$.
Case 2: $x=2$, then $y^4=100-2^4=84 \Rightarrow \emptyset$.
A: A worst-case proof by cases could consider the cases $x\ge 4$, $y\ge4$, and all nine cases of $x$ and $y$ having one of the values $1,2,3$ each.
You can save alot of work by condensing this into the cases $\max\{x,y\}\ge 4$, $\max\{x,y\}\le 2$, and $\max\{x,y\}=3$.
A: We have that 
$y^4=100-x^4=(10-x^2)(10+x^2).$ Note that $1\le x\le 3$ ($1\le x$ since we look for positive integers and $x\le 3$ because in other case $y^4=(10-x^2)(10+x^2)$ is negative.$ 
So, consider the cases $x=1,2,3.$
A: Another idea you could try is to case on the parity of $x$ and $y$. Obviously they are both even or both odd; if they are both odd, then $x^4 \equiv y^4 \equiv 1 \mod 4$, so their sum is $2 \mod 4$ and cannot be 100.
If they are both even, write $x = 2k$ and $y = 2 \ell$. Then $x^4 + y^4 = 16(k^4 + \ell^4) = 100$, implying that $100$ is divisible by $16$, a contradiction.
