Is there an integer that serves as the short leg of a primitive Pythagorean triple, the long leg of another, and the hypotenuse of a third?

Question

A primitive pythagorean triple is of the form $(a,b,c)$ where $\gcd(a,b,c) = 1$. If we list these three numbers such that $a<b<c$, is there an integer $x$ where $x$ appears in all three "slots" once?

Example: $13$ appears in the third slot $(5$-$12$-$13)$ and first $(13$-$84$-$85)$ but not in the second.

Note: $x$ must be a part of the primitive Pythagorean triple, we cannot have something like $15$-$20$-$25$

Answer 1

One can easily verify that $(221, 24420, 24421)$, $(60,221,229)$, and $(21,220,221)$ are Pythagorean triples - this is an answer.

How did I find it?

A primitive Pythagorean triple is of the form $(m^2 - n^2, 2mn, m^2 + n^2)$, with $m$ and $n$ coprime and not both odd. But there's no guarantee that the legs come out in the right order.

Since $m$ and $n$ have opposite parity, $m^2 + n^2$ must be odd (and actually of the form $4k+1$). But $2mn$ is even. So any such number will be a sum of two squares, and also a difference of two squares in two different ways, one of which has $m^2 - n^2 < 2mn$ and one of which has $m^2 - n^2 > 2mn$. Furthermore, by mod 4 considerations, in both expressions we must have $m$ odd and $n$ even.

Next, we want to figure out the number of ways to write a positive integer as a difference of squares. Each expression of an odd integer as a difference of squares corresponds to a factorization. For example consider

$$33 = 17^2 - 16^2 = (17- 16)(17 +16) = 1 \times 33$$

and

$$33 = 7^2 - 4^2 = (7-4)(7+4) = 3 \times 11$$

In fact the number of ways to write a positive odd integer as a difference of squares is exactly half its number of factors. (This is not a complete proof of this fact.)

So the number we're looking for must:

• be of form 4k+1
• have at least four factors (so it can be written as a difference of two squares in at least two ways)
• be a sum of two squares. The integers which are sums of two squares are exactly those where no prime of form $4k+3$ appears to an odd power (this is a classical result in number theory).

The easiest way to get four factors is a product of two distinct primes. We want to avoid primes of form $4k+3$, so we take $5 \times 13 = 65$. Then we have $65 = 33^2 - 32^2$ giving the triple $(65, 2112, 2113)$ and $65 = 9^2 - 4^2$ gfiving the triple $(65, 72, 97)$. So this doesn't work because $65$ falls into the "low" slot both times.

Similarly $85 = 5 \times 17$ doesn't work. You need the two primes to be relatively close together in order to get $m^2 - n^2 > 2mn$, which we need.

So look at $13 \times 17 = 221$:

• $221 = 111^2 - 110^2$ gives the triple $(221, 24420, 24421)$
• $221 = 15^2 - 2^2$ gives the triple $(60, 221, 229)$
• $221 = 10^2 + 11^2$ gives the triple $(21, 220, 221)$.

Answer 2

This is possible! For example, it occurs with $1885$:

$$1885^2 + 10428^2 = 10597^2$$

$$1692^2 + 1885^2 = 2533^2$$

$$427^2 + 1836^2 = 1885^2$$

This was found through a Python search generated and executed in ChatGPT (link).

---

Edit: Once one has found such solutions, it is possible to search for related literature more effectively. For example, see:

Fässler, Albert. "Multiple Pythagorean number triples." The American mathematical monthly 98.6 (1991): 505-517. Link.

On p. 509 part (d)(1) begins with $1105$. Checking WolframAlpha, we can use the query:

solve over Z+: a^2 + 1105^2 = c^2 && gcd(a,1105,c) = 1

to see that there are no solutions for which $1105$ is the smallest number in a primitive Pythagorean triple. The next number listed in the aforelinked is $1885$, which (as indicated above) gives a solution. This can be verified in WolframAlpha; the first result for each query (linked with the corresponding letter) is precisely the solution provided above this Edit.

For a:

solve over Z+: m^2 + 1885^2 = n^2 && gcd(m,1885,n) = 1 && 1885<m<n

For b:

solve over Z+: m^2 + 1885^2 = n^2 && gcd(m,1885,n) = 1 && m<1885<n

For c:

solve over Z+: m^2 + n^2 = 1885^2 && gcd(m,1885,n) = 1 && m<n<1885