Determine all the representations of the integer $2130797 = 17^2 \times 73 \times 101 $ as a sum of two squares.

Question

Determine all the representations of the integer $2130797 = 17^2 \times 73 \times 101 $ as a sum of two squares.

attempt: Suppose $2130797$ is of the form $n = 2^kp_1^{a_1}....p_r^{a_r}q_1^{b_1}...q_s^{b_s}$ . Where $p_1,...,p_r$ are distinct primes congruent to $1 $ modulo $4$ and $q_1,....,q_s$ are distinct primes congruent to $3$ modulo $4$. Then $n $ can be written as sum of two squares in $\mathbb{Z}$. Then the number of representations of $n $ as a sum of two squares is $4(a_1 + 1)(a_2+1)...(a_r + 1)$.

Then $2130797 = 17^2 \times 73 \times 101 = (4 + i)^2(4-i)^2(8 + 3i)(8 - 3i)(10 + i)(10 - i)$

So the number of representations of $2130797 = 4(2 + 1)(1 + 1)( 1 + 1) = 48$ .

Can someone please help me ? I don't' know how to continue. I am only able to find the number of representations. But I don't know how to start determining the different ways of representing the integer as a sum of squares.

Any feedback would help. Thank you!

Answer 1

The representations counted by $48$ are not all fundamentally distinct.

First we look at representations where each number is divisible by $17$, say $17a$ and $17b$. To find the two essentially distinct possibilities for $a$ and $b$, (i) multiply $8+3i$ by $10+i$ or (ii) multiply $8+3i$ by $10-i$. We get two numbers $a+bi$. These give the representations $(17|a|)^2+(17|b|)^2$.

Next we look at representations where the numbers are not divisible by $17$. Take $(4+i)^2$ and multiply it by $(8\pm 3i)(10\pm i)$ in all $4$ possible combinations. If a product is $x+iy$, use the representation $|x|^2+|y|^2$.

From these $6$ basic representations, you can get all representations by changing order and/or signs. The full list is not really worth writing down.

Answer 2

First note that any in any representation $$2130797=a^2+b^2$$ you can replace $a$ by $-a$; replace $b$ by $-b$; interchange $a^2$ and $b^2$. So your $48$ representations are only $6$ "really different" representations, each having $8$ "modifications". (But see the note below.)

You can find the representations by using the factorisation you have found to write $2130797$ in the form $z\overline z$. One example would be $$z=(4+i)(4-i)(8-3i)(10+i)=1411-374i$$ which gives $2130797=1411^2+374^2$, and I'm sure you can find the rest.

Comment. Note that not all the $8$ "versions" are necessarily different. For example, if we start with a square such as $9$, then two versions are $$9=3^2+0^2\quad\hbox{and}\quad 9=3^2+(-0)^2$$ which are actually the same. If we start with $$50=5^2+5^2$$ and interchange the terms, we get exactly the same expression. But since $2130797$ is neither a square nor twice a square, there is no problem in your case.