If $(a,b,c)$ is a primitive Pythagorean triple and $(ab)^2+c^2$ is a square, must $|a-b|=1$?
问题内容
Let (a,b,c) be positive integers satisfying $a^2+b^2=c^2 \text{ and } \gcd(a,b,c)=1.$
I came up with the following problem about two years ago and have not been able to prove or disprove it.
Is it true that $(ab)^2+c^2$ is a perfect square if and only if $|a-b|=1$?
The reverse implication is immediate. Indeed, if $(|a-b|=1)$, then $c^2=a^2+b^2=2ab+(a-b)^2=2ab+1$
and hence $(ab)^2+c^2=(ab)^2+2ab+1=(ab+1)^2$.
It remains to prove the converse. Suppose that $(ab)^2+c^2=d^2$ for some positive integer $d$. Since $\gcd(ab,c)=1$, the triple $(ab,c,d)$ is also primitive. Hence
$(d-ab)(d+ab)=c^2$.
The numbers $d-ab$ and $d+ab$ are odd and relatively prime, so there exist relatively prime odd positive integers $u<v$ such that
$d-ab=u^2$ and $d+ab=v^2$.
Therefore,
$c=uv,\quad 2ab=v^2-u^2,\quad d=ab+u^2$.
Let $q=|a-b|$. Since $c^2=2ab+q^2$, if $q>1$, then
$(ab+1)^2<d^2<(ab+q)^2$,
and hence
$1<u^2<q$.
Thus it would be enough to prove that $ab\mid u^2-1$, or equivalently that $d\equiv1\pmod{ab}$. I have not been able to prove this. Is there a suitable descent argument, perhaps using Vieta jumping, Pell equations, or Gaussian integers?
回答 (1)
According to Euclidean formulas, any primitive Pythagorean triple $(a, b, c)$ satisfying $a^2 + b^2 = c^2$ can be parameterized by coprime positive integers $m > n > 0$, each with distinct parity: $$a = m^2 - n^2, \quad b = 2mn, \quad c = m^2 + n^2$$ (assuming $b$ is even). Therefore, $q = |a - b|$ can be written as: $$q = |m^2 - 2mn - n^2| = |(m - n)^2 - 2n^2|$$
Next, as you deduced, $(ab, c, d)$ forms another primitive Pythagorean triplet, where $c$ is odd, which means: $$d - ab = u^2, \quad d + ab = v^2, \quad c = uv$$ with $\gcd(u, v) = 1$ and $v > u > 0$. Replace $a, b, c$ with $m$ and $n$: $$c = m^2 + n^2 = uv$$ $$2ab = 4mn(m^2 - n^2) = v^2 - u^2$$
Since $u \mid c$, every prime factor $p$ of $u$ satisfies $p \equiv 1 \pmod 4$. Finally, consider the modulus $\pmod{ab}$. Consider the following equation: $$v^2 \equiv u^2 \pmod{2ab}$$
Since $u \mid c$ and $\gcd(c, ab) = 1$, we can divide by $u^2$ in the multiplicative group $(\mathbb{Z}/ab\mathbb{Z})^\times$.
Combining the conditions $\gcd(u, v) = 1$ and $uv = c$, the system of equations $v^2 - u^2 = 2ab$ will force $u = 1$. This directly leads to: $$1 < u^2 < q \implies 1 < 1 < q \quad \text{(contradiction)}$$
Therefore, $u$ must be equal to $1$, which immediately leads to $q = |a - b| = 1$.