# Primes of the form $(2p)^{2}+1$, $p$ prime, have $h^{2}+1$ as a prime divisor?

I'm an undergraduate student and I usually ask questions here about things I'm struggling with in my academical mathematical studies, but this particular question is actually more like a curiosity.

More specifically, I'm playing around with that "famous" problem of creating a sequence that only generates prime numbers. I know that many people have tried this and it is a very hard problem, but this is part of mathematics, right? I mean, attacking hard problems as a (yet) naive student at the very least to make you realize how difficult they actually are and hopefully motivates for further study on the issue. Feel free to say that I'm wasting my time, though.

So, to the question:

I realized that the numbers $N$ of the form $N=(2p)^{2}+1$, where $p$ is prime, are often prime numbers (at least for small $p)$.

This is sort of justified by the fact that it will never by divisible by a prime $q\equiv 3\pmod 4$, since it would imply that $-1$ is a quadratic residue mod $q$. Hence, all its prime divisors are $\equiv 1\pmod 4$, and in particular, all of its divisors are of this form, since $1\cdot 1\equiv 1\pmod 4$.

Now, I thought of that famous result that says that any prime $\equiv 1\pmod 4$ can be written as a sum of two squares of integers, and hence such a number $N$ would be of the form (in prime factorization) $$N=(a^{2}+b^{2})(c^{2}+d^{2})\dots(x^{2}+y^{2}).$$

My question (for now, since I've just started these investigations), is:

Is it true that, when $N$ is not prime, one of these prime factors will always be the form $h^{2}+1$?

This is happening in my particular examples. I guess this is a hard question, and I'd appreciate any efforts, or references.

Thanks.

• I think it's always good for students like us to try to solve old problems even if we do not know the full solution because it gives us the skills to tackle new mathematical problems by ourselves which can be useful in mathematical research or when a complex problem in industry comes up that needs to be manipulated before it can be solved easily. In any case, I'm glad you asked this question because you'll interest the other people, like me, here on Math StackExchange, too, because this is an interesting problem. Hard problems like these are why I like this site so much. – Noble Mushtak Apr 18 '16 at 23:15
• Have you read about Golbachs theorem? – N.S.JOHN Apr 24 '16 at 5:24

Your hypothesis has many examples that seem to work: $$p \in \{2, 5, 7, 11, 13, 19, 29, 31, 37, 41, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 101, 103, 109, 131, 139, 149, 151, 157, 163, 179, 181, 191, 193, 197, 199\}$$ All have $(2p)^2+1$ being prime or having a prime divisor in the form of $h^2+1$ according to a Python program I just made). However, your hypothesis does not work from all prime $p$. Consider $p=17$. In this case, $(2p)^2+1=1157=13*89$, yet neither $13$ nor $89$ can be written as $h^2+1$ for $h \in \Bbb{N}$.

I wonder, though, if there is a condition for the prime $p$ in which this works.

P.S. For those interested, here is my Python code. Sorry it's not commented, I put this together very quickly using previous code I had written in order to answer this problem as quickly as I could.

Some examples where no prime factors are $h^2 + 1$

p     4 p^2 + 1
17    1157 prime fac 13 . 89 .
23    2117 prime fac 29 . 73 .
43    7397 prime fac 13 . 569 .
97    37637 prime fac 61 . 617 .
107    45797 prime fac 41 . 1117 .
113    51077 prime fac 13 . 3929 .
127    64517 prime fac 149 . 433 .
137    75077 prime fac 193 . 389 .
167    111557 prime fac 281 . 397 .
173    119717 prime fac 13 . 9209 .
227    206117 prime fac 53 . 3889 .
263    276677 prime fac 337 . 821 .
277    306917 prime fac 13 . 23609 .
283    320357 prime fac 457 . 701 .
307    376997 prime fac 277 . 1361 .
313    391877 prime fac 29 . 13513 .
347    481637 prime fac 13 . 37049 .
353    498437 prime fac 41 . 12157 .
383    586757 prime fac 29 . 20233 .
397    630437 prime fac 229 . 2753 .
433    749957 prime fac 13 . 57689 .
443    784997 prime fac 181 . 4337 .
463    857477 prime fac 61 . 14057 .
467    872357 prime fac 41 . 21277 .
487    948677 prime fac 29 . 32713 .
503    1012037 prime fac 13 . 77849 .
523    1094117 prime fac 193 . 5669 .
557    1240997 prime fac 29 . 42793 .
577    1331717 prime fac 317 . 4201 .
607    1473797 prime fac 13 . 73 . 1553 .
613    1503077 prime fac 509 . 2953 .
617    1522757 prime fac 421 . 3617 .
643    1653797 prime fac 181 . 9137 .
673    1811717 prime fac 29 . 62473 .
727    2114117 prime fac 53 . 113 . 353 .
757    2292197 prime fac 53 . 61 . 709 .
787    2477477 prime fac 97 . 25541 .
823    2709317 prime fac 13 . 208409 .
853    2910437 prime fac 73 . 39869 .
857    2937797 prime fac 1489 . 1973 .
863    2979077 prime fac 53 . 56209 .
877    3076517 prime fac 41 . 75037 .
907    3290597 prime fac 89 . 36973 .
953    3632837 prime fac 13 . 113 . 2473 .
977    3818117 prime fac 229 . 16673 .
997    3976037 prime fac 13 . 305849 .
1093    4778597 prime fac 233 . 20509 .
1097    4813637 prime fac 1721 . 2797 .
1117    4990757 prime fac 269 . 18553 .
1123    5044517 prime fac 41 . 61 . 2017 .
1153    5317637 prime fac 13 . 97 . 4217 .
1223    5982917 prime fac 1153 . 5189 .
1237    6120677 prime fac 109 . 233 . 241 .
1283    6584357 prime fac 13 . 137 . 3697 .

• Is $127$ the only mersenne prime such that if equal to $p$, it doesn’t have a prime divisor in the form $h^2 + 1$? – Feeds Feb 19 '18 at 15:40
• QEdit: No. The Mersenne Prime $2^{31} - 1 = 2147483647$ is also part of your list. Notice that $127 = 2^7 - 1$ and we also have that $7 = 2^3 - 1$ and $31 = 2^5 - 1$. This is to say, they are Mersenne Primes as well. – Feeds Feb 19 '18 at 15:49