Proof by cases to show no solutions in positive integers to the equation $x^4+y^4 = 100$

Question

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?

Answer 1

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 $\max\{x,y\}=3$.

Answer 2

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!

Answer 3

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 $x=1,2,3.$

Answer 4

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 $y^4=100-2^4=84 \Rightarrow \emptyset$.

Answer 5

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.