# Forms $apq +b = r^{n}$ where p,q,r are primes

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.

• 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. Jan 20, 2013 at 10:54
• I do not understand your question; Ockham. What questions? Where ? What past, this life past or other life´s past ? Jan 22, 2013 at 6:41

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 $n= 11$.

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...