Proving $|ad-bc|=1$ for the Product of Two Numbers Expressible as Sums of Two Squares

Question

If $p$ is an odd prime with $p \equiv 1(4)$, then it can be uniquely expressed by the sum of two squares, i.e., $p=a^2+b^2$ with $gcd(a,b)=1$.

We consider the integer u,v with form $1,2,p^k,2p^k$. The representation exists yet may not be unique. i.e. $u=a^2+b^2$, $v=c^2+d^2$, we require $gcd(a,b)=1$ and $gcd(c,d)=1$.

If given $uv=x^2+1$. Suppose $a>b$ and $c>d$. Is there any way to show that $|ad-bc|=1$?

For example: $13$ and $74$:

$13=3^2+2^2$, $74=7^2+5^2$, we have $13\times 74=962=31^2+1$ with $|ad-bc|=3\times 5-2\times 7=1$

Another example: $386$ and $17$

$386 = 2*13^2 = 17^2+7^2 = 13^2+13^3(drop)$, $29=5^2+2^2$ we have $386\times 17 = 6562 =81^2+1$ with $|ad-bc|=|17\times 2-5\times 7|=1$

Answer 1

I figure out the answer myself and post a brief proof here.

1. $1,2,p^k,2p^k$ can be represented by the sum of two squares uniquely. $u=a^2+b^2$ with $gcd(a,b)=1$

We know that the Fermat condition: if prime $p\equiv 1 \mod 4$, then $p=a^2+b^2$ is unique and $a+b i$ is a Gaussian prime. For $p^k$, we can expand the $(a+bi)^k(a-bi)^k$ and gathering the coefficients, then found out $p^k=(c+di)(c-di)$. $c+di$ can only be factorized by prime $a+bi$. Any possible $(c,d)$ is a combination of the power of term $a+bi,a-bi$. Then it's easy to see $(c,d)$ is unique for $p^k$ and $gcd(c,d)=1$. For $2p^k$ we need to multiply $(1+i)(1-i)$ and same thing apply. $2p^k$ may have $p^k+p^k$ when $k$ is even yet $gcd(p^k,p^k)=p^k$.

2. $u,v$ can only choose from $1,2,p^k,2p^k$. To satisfy the condition that $uv=x^2+1$, $uv$ can only be $1,2,p^k,2p^k,p_1^{k_1}p_2^{k_2},2p_1^{k_1}p_2^{k_2}$.

$x^2+1\equiv 1,2 \mod 4$ so $4p_1p_2$ is ruled out.

3. Now we have the Gaussian prime factorization of $uv$.

For example, we let $u*v=p_1^{k_1}p_2^{k_2}=(a+bi)^{k_1}(a-bi)^{k_1} (c+di)^{k_2}(c-di)^{k_2} =x^2+1\equiv N$. The possible solutions $(x,y)$ for $x^2+y^2=N$ depend on the factorization of N on Gaussian primes while we know the factorization is $(a+bi)^{k_1}(a-bi)^{k_1} (c+di)^{k_2}(c-di)^{k_2}$. Since $gcd(a,b)=gcd(c,d)=1$, it's not possible for $(a+bi)(c+di)$ or $(a+bi)(c-di)$ to be pure integer. So we only have the possible solution for N:

4. $(a^2+b^2)(c^2+d^2)=uv=x^2+1^2=N$, from which we have the Brahmagupta Identity:

$(a^2+b^2)(c^2+d^2) = (ac+bd)^2+ (ad-bc)^2 = (ad+bc)^2+(ac-bd)^2$

So only two possible solutions for $x^2+y^2=N$:

$(ac+bd,ad-bc)$ and $(ad+bc,ac-bd)$

If we assume $a>b,c>d$, then $ac-bd>ad-bc$, thus we have $|ad-bc|=1$.