Problems & Puzzles: Puzzles

Puzzle 184. Gronau's prime triplets

Daniel Gronau sent (16/6/2002) the following puzzle:

Find triples (p,q,r) of primes, p<q<r, such that:

p*q+r = x²,
p*r+q = y²
q*r+p = z²

As a mater of fact he has already found some solutions:

(3,577,10369)
(17,19,2593)
(577,881,1459)
(2591,32257,34849)
(3041,16561,19603)
(4049,126001,130051)
(7687,13121,20809)
(46817,65521,112339)

He also has the following question:

1) Is there a triplet for every prime p?

I would like to add some more questions:

2) Can you find an example of two solutions for the same p?

3) Is there a limit for the number of solutions for a fixed p value?

4) Do you devise a strategy to find the asked solutions by Gronau

5) Is this problem connected to another?

Solution:

Luis Rodríguez found one example for the question 2:

(3, 10369, 186049)

Gronau also sent the following two solutions related to this question:

(17, 2593, 368083)
(19, 2593, 378449)

and one observation:

If p,q,r > 3 then 6|x, 6|y, 6|z.

***

J. C. Rosa solved (3/7/2002) the main and first question of this puzzle:

About the question 1 of the Puzzle 184 , here is my answer:

Conclusion:

If p,q,r >3 , the Gronau's prime triplets only exist
if p,q,r are of the form : 18*k+/-1

Here is the proof :

1) It is obvious that a square is equal to 0,1,3 or 4 mod 6
2) If p=1 mod 6 then we have only two solutions for q and r
(else we obtain a square equal to 2 mod 6:

a) q=1 mod 6 and r=-1 mod 6
Thus p*q+r=0 mod 6 and therefore x=0 mod 6 and
x^2=0 mod 6.
Let p=6*a+1,q=6*b+1,r=6*c-1 , x^2=36*k1^2 ,
y^2=36*k2^2,z^2=36*k3^2
After some calculations the following equalities:
p*q+r=x^2
p*r+q=y^2
q*r+p=z^2
become:
6*a*b+a+b+c=6*k1^2
6*a*c-a+b+c=6*k2^2
6*b*c+a-b+c=6*k3^2
from where:
a+b+c=0 mod 6
-a+b+c=0 mod 6
a-b+c=0 mod 6
and consequently: a=b=c=0 mod 3

therefore we obtain the triplet:
p=18*n+1,q=18*m+1,r=18*s-1

b) q=-1 mod 6 and r=1 mod 6

With the same notations as above we obtain:

6*a*b-a+b+c=6*k1^2
6*a*c+a+b+c=6*k2^2
6*b*c+a+b-c=6*k3^2
from where:
-a+b+c=0 mod 6
a+b+c=0 mod 6
a+b-c=0 mod 6

and consequently: a=b=c=0 mod 3

therefore we obtain the triplet:
p=18*n+1,q=18*m-1,r=18*s+1

3) if p=-1 mod 6 then also we have only two solutions for q and r
and by the same way as above we obtain:
a) a=b=c=0 mod 3 and the triplet:

p=18*n-1 ,q=18*m+1 ,r=18*s+1

b) a=b=c=0 mod 3 and the triplet:

p=18*n-1,q=18*m-1,r=18*s-1

C.Q.F.D (in
French : Ce Qu'il Fallait Démontrer )

Consequently for the question 1 the answer is :
none triplet if p
= 5;7;11;13;23;29;.....etc

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