# 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?

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 $$a, is there an integer $$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 $$15$$-$$20$$-$$25$$

• Didn't check my answer well enough but (65,72,97),(33,65,73),(16,63,65) is a close one Commented Aug 14 at 14:25
• Well I do know that x is odd, so that narrows my search down by 50% Commented Aug 14 at 14:51
• @uggupuggu Also , $x$ must be of the form $4k+1$ and not divisible by $3$. Commented Aug 14 at 14:55
• @Peter - Indeed, $x$ must not be divisible by any $4k+3$ (by the known properties of Gaussian primes, and the fact that the sought Gaussian integer must not have ordinary integer factors). Commented Aug 15 at 2:56
• According to this snippet I wrote, the least such $x$ if you drop the gcd condition is $x=15$, viz. the triples $(15,\,20,\,17)$, $(8,\,15,\,17)$ & $(9,\,12,\,15)$. If you demand all three triples are co-prime as per this snippet, you get @MichaelLugo's solution with $x=221$ as the minimal $x$.
– J.G.
Commented Aug 15 at 10:32

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 $$(21, 220, 221)$$.
• very nice!! this is increasingly looking to me like a fun homework problem for the upcoming year Commented Aug 14 at 16:08
• Now , my routine worked better. For at least the $20$ following numbers there are solutions using $a,b\le 400$ : $$[221, 325, 377, 425, 493, 629, 697, 725, 1885, 2405, 2665, 3145, 3445, 3485, 3965, 4505, 4745, 5185, 5785, 6205]$$ Commented Aug 14 at 16:14
• @BenjaminDickman I enjoyed figuring it out - I don't know if your students would! Commented Aug 14 at 16:51
• Also $221=5^2+14^2$ gives the triple $(140,171,221)$. Commented Aug 14 at 21:24
• @Peter You might consider adding the sequence of integers from your comment yesterday to the On-Line Encyclopedia of Integer Sequences! It doesn't seem to be there yet. Commented Aug 15 at 16:24

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