Writing $28913000$ as the sum of two squares

Question

A little number theory fun. I am given that $167^2 + 32^2 = 28913$, and I am asked to find integers $a$ and $b$, such that $a^2 + b^2 = 28913000$.

Here's my thought process so far:

Knowing that $1000 = 10^2 + 30^2$, I rewrote $28913000$ as $28913\times 1000$, and proceeded to multiply the sums of squares:

$$(167^2 + 32^2)(10^2 + 30^2).$$

However, after foiling, I ended up with the sum of $4$ squares, and cannot think of a way to just find two squares, $a$ and $b$.

Any help would be greatly appreciated!

Answer 1

Very good!

Do you know complex numbers? Assume that $-1$ has a square root somewhere (certainly not in $\mathbb R$), denote it $i$, and introduce $+$, $\cdot$ operations with reals and $i$. So, $i^2=-1$, thus $(a+bi)(a-bi) = a^2+b^2$. $$(a+bi)(a-bi)(c+di)(c-di) = (a+bi)(c+di)\cdot (a-bi)(c-di)$$ Can you calculate it?

Answer 2

You can easily check that $$ (a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2 $$

Answer 3

One can also find all of the solutions computationally. E.g. here's some GAP code:

n:=28913000;;

for a in [0..RootInt(n)] do
  b:=n-a^2;
  c:=RootInt(b);
  if(IsSquareInt(b)) then
    Print(a^2+c^2," = ",a,"^2 + ",c,"^2\n");
  fi;
od;

which gives

28913000 = 182^2 + 5374^2
28913000 = 710^2 + 5330^2
28913000 = 1330^2 + 5210^2
28913000 = 2062^2 + 4966^2
28913000 = 2174^2 + 4918^2
28913000 = 2630^2 + 4690^2
28913000 = 3370^2 + 4190^2
28913000 = 3766^2 + 3838^2
28913000 = 3838^2 + 3766^2
28913000 = 4190^2 + 3370^2
28913000 = 4690^2 + 2630^2
28913000 = 4918^2 + 2174^2
28913000 = 4966^2 + 2062^2
28913000 = 5210^2 + 1330^2
28913000 = 5330^2 + 710^2
28913000 = 5374^2 + 182^2