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 6x, 6y, 6z.
***
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*c1 , 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*ca+b+c=6*k2^2
6*b*c+ab+c=6*k3^2
from where:
a+b+c=0 mod 6
a+b+c=0 mod 6
ab+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*s1
b) q=1 mod 6 and r=1 mod 6
With the same notations as above we obtain:
6*a*ba+b+c=6*k1^2
6*a*c+a+b+c=6*k2^2
6*b*c+a+bc=6*k3^2
from where:
a+b+c=0 mod 6
a+b+c=0 mod 6
a+bc=0 mod 6
and consequently: a=b=c=0 mod 3
therefore we obtain the triplet:
p=18*n+1,q=18*m1,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*n1 ,q=18*m+1 ,r=18*s+1
b) a=b=c=0 mod 3 and the triplet:
p=18*n1,q=18*m1,r=18*s1
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
***
