If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.

Question

If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.

Came across this question while solving an exercise on prime numbers.

Now I found a pretty simple solution to this problem here.

(It is not the exact same problem but the method described there in most of the answers can be easily used on this one too.)

But when I was trying to solve this on my own, I took another path.

I noticed that $a^2 + b^2 + c^2 + d^2$ is even (and therefore composite) if:

(1) All of $a, b, c, d$ have the same parity (i.e. all are odd or all are even).

(2) Two of them are even and two are odd (say $a, c$ are even and $b, d$ are odd).

So the only case left to worry about is when $a, b, c$ are even and $d$ is odd (wlog).

Example: $26\times 28 = 56\times 13$ ...and... $13^2 + 26^2 + 28^2 + 56^2 = 4765$.

And this is where I'm stuck. Any ideas?

EDIT: As pointed out in some if the answers, my initial assumption of the given expression always being divisible by 5 was wrong (I apologise for that). But I'd still like to see if it can proved, in the manner I started out, that the given expression is composite.

@Answerers Please don't post here proofs different from the one the OP proposes. Most common proofs are already in the link given in the OP. If you know a new proof then please post it in that linked post - not here. — Bill Dubuque, Jan 09 '22 at 13:31
$1 \times 0 = 0 \times 1$ and $1^2 + 0^2 + 0^2 + 1^2 = 2$ is prime. — Essaidi, Jan 09 '22 at 15:42
Observe: with $ab=cd$ the cross terms in $(a+b)^2+(c-d)^2$ and $(a-b)^2+(c+d)^2$ cancel, so both of these two-square sums equal $a^2+b^2+c^2+d^2$. — Oscar Lanzi, Jan 09 '22 at 16:57

score 4 · Answer 1 · answered Jan 09 '22 at 14:35

4

It's not always divisible by $5$:

$6 \times 6 = 4 \times 9$ but $6^2+6^2+4^2+9^2 = 169$

answered Jan 09 '22 at 14:35

jjagmath

18,214

Thanks for that. I also had a feeling about it, but I tried about 10 different cases and I got a multiple of 5 every single time. It was a mistake to assume what I did. Thanks for a counter example, but I'd still like to know if there's a way of proving the initial question in the way I started (now that divisibility by 5 is out of the way). – Nick Larry Jan 10 '22 at 13:08
I think it's unlikely that you found a proof looking at divisibility by a fixed set of primes. Consider for example: $4^2+6^2+24^2+1^2 = 17\times 37$, $6^2+10^2+4^2+15^2 = 13\times 29$, $10^2+28^2+8^2+35^2 = 41\times 53$, etc... – jjagmath Jan 10 '22 at 14:15

If $a, b, c, d$ are natural numbers, such that, $ab = cd$, prove that $a^2 + b^2 + c^2 + d^2$ is a composite number.

1 Answers1