Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Some small results for $2pq +3 = r^{n} $ p,q,r primes; written in the form (p,q,r,n):

$(3,1093,3,8) (59,997,7,6) (73,107,5,6) (7,223,5,5) (3,13,3,4) (11,109,7,4) (109,131,13,4) (277,1667,31,4) (5,491,17,3) (89,137,29,3) (11,13,17,2)$

Some small results for $2pq +1 = r^{n} $ p,q,r primes; written in the form (p,q,r,n):

$(13,757,3,9) (2,19531,5,7) (11,11,3,5) (3,2801,7,5) (29,3541,59,3) (2,31,5,3) (2,2,3,2)$ Last one is in fact $2^{3}+ 1=3^{2}$

I do not understand for now, though, why the first form $ 2pq + 3$ produces much more powers than the second. The form $ 3pq + 2$ produces (up to primes < $10^{5}$) releasing the condition for r to be a prime: 1 9-th power; 1 7-th power; 4 5-th powers, 38 cubes, but 0 squares as the form is always 2 mod 3 and squares are 0 or 1 mod 3.

share|improve this question
    
N.S. The range i checked for n powers was too low (n<=9) to draw conclusions.The powers from 2 to 9 are all primes with the exception of 9 and 4,6, and 8 which are squares and the form 2pq +1 allows only a square: (2,2,3,2) though i still do not see why. –  user55514 Jan 20 '13 at 10:54
    
Why haven't you accepted any answers in the past? –  Math_Illiterate Jan 21 '13 at 15:22
    
I do not understand your question; Ockham. What questions? Where ? What past, this life past or other life´s past ? –  user55514 Jan 22 '13 at 6:41
    
Haven't you noticed underneath your name it says 0% accept rate, that means you haven't clicked the green check mark next to anyone who has helped you in the past with problems. When you click that button it gives the person increased reputation points and it's like a "thank you for helping me with my problem" sort of thing. You should accept peoples answers if they help you out at all. –  Math_Illiterate Jan 22 '13 at 18:40

2 Answers 2

Not a "mathematical" answer, but I would expect the first one to produce mosr solutions for the following simple reason:

If $d|n$ then $r^d-1|r^n-1=2pq$.

Since $p,q$ are prime, then either $r=3$ and $n$ has at most $2$ divisors, or $r>3$ and $n$ must be prime. This reduces a lot the range of potential solutions.

There seems to be no similar constrain on $2pq+3=r^n$, it is easy to find couple constrains, but none as big as the above...

Anyhow, keep in mind that is purely a heuristic argument, similar arguments can be used to draw wrong conclusions...

share|improve this answer

Primes are $1$ or $3 \mod 4$, else they would be divisible by $2$ or by $4$. Squares are always $0$ or $1 \mod 4$ since $2^2=4 = 0 \mod 4$ and $3^2=9=1 \mod 4$ and so if $a = 2\text{ or }3 \mod 4$, then $a^2= 0\text{ or }1 \mod 4$ and if $a=0\text{ or }1 \mod 4$ then $a^2= 0\text{ or }1 \mod 4$.

1) If $p$ and $q$ are both $1 \mod 4$, then $pq = 1 \mod 4$ and $2pq +1 = 3 \mod 4$ which is never a square.

2) If $p$ and $q$ are both $3 \mod 4$, then $pq = 1 \mod 4$ and $2pq +1 = 3 \mod 4$ which cannot be a square.

3) If one of $p$ and $q$ is $1 \mod 4$ and the other is $3 \mod 4$, then $pq = 3 \mod 4$ and $2pq +1 = 7 \mod 4 = 3 \mod 4$ which cannot be a square.

So no matter what are the primes, the form $2pq +1$ is never a square but when $p=q=2$ and $2pq+1 = 9 \mod 4 = 1 \mod 4 \implies 2 \cdot 2 \cdot 2 +1 = 2^3 +1 = 2^2 + 2^2 +1 = (2 +1)^2 = 3^2$.

I checked first $\mod 3$, but you cannot derive any conclusion. Fortunately $4 = 3+1$ and the intellectual work and effort to do was short.

There is still left the case $p=2$ and $q$ any odd prime. The form is then $4q +1$ and cannot be a square because odd primes are of the form $2k +1 \implies 4q +1 = 8k + 5$ and the squares are $0, 1\text{ or }4 \mod 8 \implies 4q + 1$ is never a square.

But if the form $2pq +1$ does not admit squares but for $p=q=2$, it will not admit any even power which are also squares; thus limiting the number of powers $2pq+1 = a^n$ this form admits.

Forms $9pq +3$ and $9pq +6$ do not admit powers $a^n$ up at least to $n= 11$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.