# New property of primes

## Main Question or Discussion Point

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

coolul007
Gold Member
$a^{p-1} - 1 \equiv0 mod(p)$ the factorization is: $(a^{\stackrel{p-1}{2}} - 1)(a^{\stackrel{p-1}{2}} + 1)$

in (a), doesn't the RHS reduce to (P+1)/2, allowing that to be cancelled 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 in to 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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$a^{p-1} - 1 \equiv0 mod(p)$ 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

haruspex
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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).

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

haruspex
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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.

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.

haruspex
Homework Helper
Gold Member
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.

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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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.