# Distribution of primes in primitive Pythagorean triples

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


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.

• 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 Commented Apr 20 at 23:47
• @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? Commented Apr 21 at 0:34
• @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 Commented Apr 21 at 0:41
• 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. Commented Apr 21 at 14:25
• @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 Commented Apr 21 at 17:28