# What is the mysterious identity that holds for primes and certain composites?

• ramsey2879
In summary: I have not found a proof or a reference to it. For P prime it is related to the fact that the sum of the quadratic residues and the sum of the quadratic nonresidues mod P are both -1 mod P. This is true for all primes not just those of the form 8n+1 or 8n+7.In summary, for a prime number P mod 8, the following identities hold: a) \frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\
ramsey2879
I would like to show that if a prime number P mod 8 is a) 1 or 7 or b) 3 or 5 then

a) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} + 2)$$ mod P

b) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} - 2)$$ mod P

Any suggestions

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in (a), doesn't the RHS reduce to (P+1)/2, allowing that to be canceled as a factor throughout?
This doesn't apply to (b), so there's a curious asymmetry.

$a^{p-1} - 1 \equiv0 mod(p)$ the factorization is: $(a^{\stackrel{p-1}{2}} - 1)(a^{\stackrel{p-1}{2}} + 1)$

haruspex said:
in (a), doesn't the RHS reduce to (P+1)/2, allowing that to be canceled as a factor throughout?
This doesn't apply to (b), so there's a curious asymmetry.
That is correct. The formula on the left hand side is the direct formula for S_((P-1)/2) where S_(0) = (P+1)/2; S_(1) = - S_(0) and S_(n) = 6S_(n-1) - S_(n-2). In a) the factor (P+1)/2 does cancel out but I left it into show the relationship to b). There is a curious asymmetry for most all recursive series of the form S_n = 6S_(n-1) - S_(n-2) mod P depending upon whether P is a prime of the form 8n +/- 1 or 8n +/-3, but I agree that the one here is quite curious.

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coolul007 said:
$a^{p-1} - 1 \equiv0 mod(p)$ the factorization is: $(a^{\stackrel{p-1}{2}} - 1)(a^{\stackrel{p-1}{2}} + 1)$
Although that is true for integer a, I don't see that it applies to (3 +/- sqrt8)^(P-1)/2

Isn't there a factor of 2 missing somewhere in (a)? After cancelling the (P+1)/2, the LHS always produces an even number. But it does appear to be always 2 when P congruent to 1 or 7 (8).

ramsey2879 said:
I would like to show that if a prime number P mod 8 is a) 1 or 7 or b) 3 or 5 then

a) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} + 2)$$ mod P

b) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} - 2)$$ mod P

Any suggestions

What you have in (a), after a little algebra, is
$$\frac{p+1}{2}\left(1-\sqrt 2\right)\left(3+\sqrt 8\right)^{(p-1)/2}+\frac{p+1}{2}\left(1+\sqrt 2\right)\left(3-\sqrt 8\right)^{(p-1)/2}=\frac{p+1}{2}\Longleftrightarrow (1-\sqrt 2 )(3+2\sqrt 2)^{(p-1)/2}+(1+\sqrt 2)(3-2\sqrt 2)^{(p-1)/2}=1$$

But, for example for $\,p=7\,$ , we get:
(3\pm2\sqrt 2)^3=27\pm 54\sqrt 2+72\pm 16\sqrt 2\Longrightarrow \begin{align*} (1-\sqrt 2)(3+2\sqrt 2)^3=&-41-29\sqrt 2\\(1+\sqrt 2)(3-2\sqrt 2)^3=&-41+29\sqrt 2\end{align*}

If we add the above we get $\,-82\neq 1\,$ , so either I'm making a mistake, or I didn't understand correctly...or the formula's wrong.

DonAntonio

DonAntonio said:
If we add the above we get $\,-82\neq 1\,$ , so either I'm making a mistake, or I didn't understand correctly...or the formula's wrong.
DonAntonio
As I pointed out in post #6, there seems to be a factor of 2 error.
If you halve the LHS you get -41, which is 1 mod 7 as required.

haruspex said:
As I pointed out in post #6, there seems to be a factor of 2 error.
If you halve the LHS you get -41, which is 1 mod 7 as required.

Your right. There is a factor of 2 error in both a) and b). As I said in a later post. This is a direct formula for S_(n) where S_(0) = (P+1)/2 and S_(1) = - S_(0). I am sorry for the error. If you substitute (P+1)/4 for (P+1)/2 you get correct results for S_(0), S_(1) and
S_(3).

Thankyou haruspex and DonAntonio for pointing out my error so it could be corrected.

Fwiw, I investigated the behaviour of sk+1 = 2 sk + sk-1 mod p, s0=s1=1, looking for where it hits 1. I see the same dependency on p mod 8:
- for 1 or 7 mod 8, it repeats every p-1 terms
- for 3 or 5 mod 8, it repeats every 2p+2 terms
There are relatively few where it repeats more often as well. It repeats X times as often as follows:
p X
29 3
41 4
59 3
79 3
103 3
137 2
179 5
197 11
199 11
:
Strange sequence. Also looked at modulo odd composites up to 91. No obvious patterns. The only one that behaved like a prime (i.e. as above) was 35.

haruspex said:
Fwiw, I investigated the behaviour of sk+1 = 2 sk + sk-1 mod p, s0=s1=1, looking for where it hits 1. I see the same dependency on p mod 8:
- for 1 or 7 mod 8, it repeats every p-1 terms
- for 3 or 5 mod 8, it repeats every 2p+2 terms
There are relatively few where it repeats more often as well. It repeats X times as often as follows:
p X
29 3
41 4
59 3
79 3
103 3
137 2
179 5
197 11
199 11
:
Strange sequence. Also looked at modulo odd composites up to 91. No obvious patterns. The only one that behaved like a prime (i.e. as above) was 35.
This sequence can be broken up into two intermeshed sequences of the form Sk+1 = 6*Sk - Sk-1. Is it correct that you were aware of this? PS see OEIS sequences A182431, A182439 through A182441 which I posted. The relation to the the sequences here where S0 = N and S1=-N (same for your sequence type and my sequence type) is explained in the comments and in the sequences referenced there.

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ramsey2879 said:
I would like to show that if a prime number P mod 8 is a) 1 or 7 or b) 3 or 5 then

a) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} + 2)$$ mod P

b) $$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P-1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P-1}{2} = (\frac{P-3}{2} - 2)$$ mod P

Any suggestions
I also have found that the following identity holds and it does not depend on the value of P mod 8:

$$\frac{(P+1)}{2}(1-sqrt{2})(3+sqrt{8})^\frac{P+1}{2}+ \frac{(P+1)}{2}(1+sqrt{2})(3-sqrt{8})^\frac{P+1}{2} = -1$$ mod P

It appears to hold for all primes and certain composites such as 169.

## 1. What is a "new property of primes"?

A "new property of primes" refers to a recently discovered pattern or characteristic that applies to prime numbers. Prime numbers are positive integers that are only divisible by 1 and themselves. They play a fundamental role in number theory and have been extensively studied for centuries.

## 2. How are these new properties of primes discovered?

New properties of primes are typically discovered through mathematical research and analysis. This involves exploring patterns, relationships, and equations involving prime numbers. With the advancement of technology, computer algorithms have also been used to assist in the discovery of new properties of primes.

## 3. Can you provide an example of a new property of primes?

One example of a new property of primes is the Green-Tao theorem, which states that there are arbitrarily long arithmetic progressions of primes. This means that there are infinitely many sets of prime numbers that have a constant difference between them, such as 3, 7, 11, 15, 19, where the difference between each number is 4.

## 4. How do new properties of primes impact mathematics?

New properties of primes can have a significant impact on mathematics as they provide deeper insights and understanding into the behavior of prime numbers. They can also lead to the development of new theories and applications in various fields such as cryptography, number theory, and computer science.

## 5. Are there still many unknown properties of primes?

Yes, there are still many unknown properties of primes. Despite centuries of research, prime numbers continue to hold many mysteries and their behavior is not fully understood. The discovery of new properties of primes is an ongoing and active area of research in mathematics.

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