Problem 2064

We show by induction on $n \ge 0$, that $2^{n+1}$ appears in precisely $n$ distinct Pythagorean triples. Since no Pythagorean triple contains $2$, the assertion holds for $n=0$.

For the induction step, let $n \ge 1$, and assume that $2^n$ appears in exactly $n-1$ distinct Pythagorean triples. The latter produce $n-1$ distinct non-primitive Pythagorean triples each containing $2^{n+1}$. To conclude the proof, we show that $2^{n+1}$ appears exactly once in a primitive Pythagorean triple. Recall that the primitive Pythagorean triples are described by the well-known formulae $x = v^2 - u^2$, $y = 2uv$, $z = u^2 + v^2$, where $u$ and $v$ are coprime positive integers, not both odd, and $u < v$. Since $x$ and $z$ are both odd, if $2^{n+1}$ appears in the triple, then $2^{n+1} = y = 2uv$, and since $u < v$ and $u$ and $v$ have opposite parity, necessarily $u=1$ and $v = 2^n$. Consequently, $2^{n+1}$ appears in exactly $n$ distinct Pythagorean triples.

Alternative solution 1:
If $P(m)$ is the number of Pythagorean triples containing the positive integer $m$, and if $P_0(m)$ is the number of primitive such triples, then $P(m) = \sum_{d|m} P_0(d)$. Since $P_0(1) = P_0(2) = 0$ and $P_0(2^k) = 1$, $k \ge 2$ (as in the previous solution), it follows that $P(2^{n+1}) = n$, so $2^{n+1}$ appears in exactly $n$ distinct Pythagorean triples.

Alternative solution 2:
We show that if $p$ is a prime congruent to $3$ modulo $4$, then $p^n$ appears in exactly $n$ Pythagorean triples, and is moreover always the smallest entry of any such.
Since $p$ is congruent to $3$ modulo $4$, $-1$ is a quadratic non-residue modulo $p$, so no power of $p$ can be the largest entry of a Pythagorean triple. Hence, if $p^n$ is a member of a Pythagorean triple, then $p^{2n} = b^2 - a^2$ for some positive integers $a < b$, so $b - a = p^k$ and $b + a = p^{2n-k}$ for some non-negative integer $k < n$. Clearly, every such $k$ corresponds to a solution and there are precisely $n$ distinct Pythagorean triples containing $p^n$, namely,
$$p^n, \quad p^k(p^{2(n-k)} - 1)/2, \quad p^k(p^{2(n-k)} + 1)/2, \quad k = 0, 1, \dots, n-1.$$
It is worth noticing that this argument avoids appealing to the parametric representation of Pythagorean triples.