My Observation:

I've observed a pattern where for every pair of twin primes ($p$, $p+2$), there appears to be at least one primitive Pythagorean triple ($a$, $b$, $c$) such that one of the twin primes is equal to the hypotenuse ($c$) of the triple. Here, primitive means that $gcd(a, b, c) = 1$

Exploration and Verification:

To investigate this further, I've been utilizing Euclid's formula for generating primitive Pythagorean triples:

$a = m^2 - n^2$
$b = 2mn$
$c = m^2 + n^2$

where $m$ and $n$ are coprime integers with $m > n$. I've been substituting suitable values of $m$ and $n$ to derive triples $(a, b, c)$ and verifying that $c$ is indeed one of the twin primes in the pair $(p, p+2)$. I've compiled a list of confirming examples of twin prime pairs less than 1021 and associated primitive Pythagorean triple ;

highest twin prime confirmed with associated primitive Pythagorean triple (12415229, 12415231) $m=2498$, $n=2485$, $a=64779$, $b=12415060$, $c=12415229$

Seeking Assistance:

However, I'm seeking assistance from the mathematical community to:

Formalize my conjecture: Can we turn this pattern into a rigorous conjecture or theorem about the distribution of primes within primitive Pythagorean triples? Explore implications: Are there any interesting number-theoretic properties to explore based on this connection? While it might not directly prove the twin prime conjecture, could it shed light on prime distribution patterns in a way that hasn't been explored before?

Has anyone found any counterexamples?

Is there a known twin prime pair $(p, p+2)$ for which there exists no primitive Pythagorean triple $(a, b, c)$ where c is equal to one of the primes?

note this is every pair less than 1021

(3, 5) m=2, n=1, a=3, b=4, c=5
(5, 7)  m=2, n=1, a=3, b=4, c=5
(11, 13)  m=3, n=2, a=5, b=12, c=13
(17, 19) m=4, n=1, a=15, b=8, c=17
(29, 31) m=5, n=2, a=21, b=20, c=29
(41, 43) m=5, n=4, a=9, b=40, c=41
(59, 61) m=6, n=5, a=11, b=60, c=61
(71, 73)  m=8, n=3, a=55, b=48, c=73
(101, 103) m=10, n=1, a=99, b=20, c=101
(107, 109) m=10, n=3, a=91, b=60, c=109 
(137, 139) m=11, n=4, a=105, b=88, c=137
(149, 151) m=10, n=7, a=51, b=140, c=149
(179, 181) m=10, n=9, a=19, b=180, c=181
(191, 193) m=12, n=7, a=95, b=168, c=193
(197, 199) m=14, n=1, a=195, b=28, c=197
(227, 229) m=15, n=2, a=221, b=60, c=229
(239, 241) m=15, n=4, a=209, b=120, c=241
(269, 271)m=13, n=10, a=69, b=260, c=269
(281, 283) m=16, n=5, a=231, b=160, c=281
(311, 313) m=13, n=12, a=25, b=312, c=313
(347, 349) m=18, n=5, a=299, b=180, c=349
(419, 421) m=15, n=14, a=29, b=420, c=421
(431, 433) m=17, n=12, a=145, b=408, c=433
(461, 463) m=19, n=10, a=261, b=380, c=461
(521, 523) m=20, n=11, a=279, b=440, c=521
(569, 571) m=20, n=13, a=231, b=520, c=569
(599, 601) m=24, n=5, a=551, b=240, c=601
(617, 619) m=19, n=16, a=105, b=608, c=617
(641, 643) m=25, n=4, a=609, b=200, c=641
(659, 661) m=25, n=6, a=589, b=300, c=661
(809, 811) m=28, n=5, a=759, b=280, c=809
(821, 823) m=25, n=14, a=429, b=700, c=821
(827, 829) m=27, n=10, a=629, b=540, c=829
(857, 859) m=29, n=4, a=825, b=232, c=857
(881, 883)  m=25, n=16, a=369, b=800, c=881
(1019, 1021) m=30, n=11, a=779, b=660, c=1021

highest twin prime confirmed with associated primitive Pythagorean triple 
(12415229, 12415231) m=2498, n=2485, a=64779, b=12415060, c=12415229

2 Answers 2


For every pair of twin primes $(p,p+2)$, either $p$ or $p+2$ will leave remainder $1$ when divided by $4$ (due to basic modular arithmetic).

By Fermat's theorem on the sum of two squares, every such prime is a sum of two squares. Combining this with Euclid's formula yields that your conjecture is true.


Maybe I am reading the problem wrongly, but to me there is a simple solution:

  • Every twin prime pair includes a prine $p\equiv1\bmod 4$.

  • This prime $p$ has the form $m^2+n^2$ for some natural numbers $m,n$ having opposite parity.

  • Therefore from the standard form for the gwneral primitive pythagorean triple, $p$ as indicated above is the hypotenuse of such a triple.

  • No twin prime status is really needed. Every prime $\equiv1\bmod 4$, regardless of the status of its odd neighbors, is the hypotenuse of a primitive Pythagorean triple.

  • $\begingroup$ It's my fault I may have formatted the question wrong . But by establishing that every prime pair has included a prime congruent to 1 mod 4 and by being the hypotenuse of a primitive pythagorean triple , can lead to a simple proof for twin prime conjecture since the distribution is already known for pythagorean triples $\endgroup$ Commented Apr 20 at 23:47
  • $\begingroup$ @NicholasJoseph Since there are many primitive Pythagorean triples that don't correspond to twin primes, how do you relate the distribution of twin primes to the distribution of primitive Pythagorean triples? $\endgroup$ Commented Apr 21 at 0:34
  • $\begingroup$ @StevenClark Euler found in 1747 that a prime of 1 mod 4 , then it is the hypotenuse of a Pythagorean triple doesn't necessarily address twin primes maybe this is my observation since one of prime pair has to be 1 mod 4 however I have yet to find a twin prime that has no associated PPT and yes there is many PPT that don't correlate to a twin prime however every prime pair has at least one that does $\endgroup$ Commented Apr 21 at 0:41
  • $\begingroup$ Wolfram Alpha shows primes here and OEIS shows hypotenuse values here Between them we can see, for example that these values are not in the set of twin primes. $\,53,89,97,109,113\,$ It may not affect your conjecture but it just shows there are other prime hypotenuse values. $\endgroup$
    – poetasis
    Commented Apr 21 at 14:25
  • $\begingroup$ @poetasis all 1 mod 4 primes and appear in primitive triples , (107,109 ) 109 does have a twin , I have noted a good many 1 mod 4 primes that have no twin. I feel like when I have enough data I can discern a pattern from generated values for (m,n) from Euclids but for now I'm just generating a good many pairs $\endgroup$ Commented Apr 21 at 17:28

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