Both weren't able to find more solutions than the reported one going up to
21940132061 and to 31*10^9, in their respectively search.
I tried to find solutions to:
with p,q,r,s consecutive primes.
The last two equations are added because of a remark of Farideh in
At first I decided to use a Sieve. I didnít find another solution
So time for a different attack. And I followed the suggestion by
Carlos in Puzzle 498:
where a=q-p, b=r-p, c=s-p, the prime gaps of q,r,s relative to p, so
From this we have that [0,a,b,c] mod 3 must be admissible, it may
not form a complete residue set mod 3.
Multiply by 2:
We know that 2|(a+b+c) and that 4|bc, so the expression can be
divided by 4 and y must be even y=2z.
Suppose that 4|(a+b+c). We can than complete the square:
The right hand side is even and the left hand side is odd.
So we must have that a+b+c=2 mod 4. Letís complete the square on (*)
Or after some manipulations:
g = (a+b+c)/2
d = 2bc-g^2
x = 2p+g
This diophantine equation (**) is well known. I would recommend
Diophantine Equation x^2-Dy^2=N, D>0 by Keith Matthews.
In it he describes the method of simple continued fractions to solve
this type of equation, without the necessity of using the
fundamental solution of Pellís equation (d=+/-1) in order to
describe all solutions.
In our case, a necessary condition for solutions to exist is x^2=2
So this restricts the possibilities of d to those d for which 2 is a
square residue mod abs(d).
If |d| is prime the Lengendre symbol (2/|d|) should equal 1 to allow
If |d| is not prime the Jacobi symbol (2/|d|) excludes cases if the
result equals -1.
The method requires to find all roots r in (0..|d|-1) such that r^2
= 2 mod |d|.
[If d=+/-1 one could use the fundamental solution to the resulting
So if the Jacobi symbol returns 1, it is not conclusive, one should
just try to find roots r, we need them anyway.
The method of continued fractions is used on (r+sqrt(2))/|d| for all
roots r in a reduced residue set mod |d|.
The other two equation types (II,III) are similar but have a
II) swap a-c
III) swap a-b
I searched all [a,b,c] with c<= 500, using the conditions on a,b,c
as stated before.
For these I computed a list of values of d (having 2 as a square
mod |d|) and for each d I kept a list of values of g and the
equation type that generated this combination (d,g). For every root
r belonging to a value of d I tried to find solutions x such that:
p=(x-g)/2 is prime and compute a,b,c belonging to this p and itís
realised values g* and d*. [For every possible g]
Finaly check g* and d* against the values (g,d) used to generate
this solution (x,p), using the correct type of equation belonging to
I found 16589 different d. The total number of combinations (d,g) is
equal to 572122.
I used a maximum of 105 iterates. (Limiting the primes p to
approximately 40 digits). However I found no other solutions.
As an example:
For the given solution 203233, we have (a=16,b=46,c=60), so d=1799
Within my search bounds for [a,b,c] there are 132 values of g, one
of which is 61.
1799 has 4 roots r (60,711,1088,1739) with r^2=2 mod 1799.
The root 711 defines the number (711+sqrt(2))/1799.
The associated simple periodic continued fraction is [0; 2 1 1 9 2],
the last 2 repeats indefinitely.
Its value as a fraction terminated at iterate 14 is equal to
So x = 113835*1799 Ė 287448*711 = 406527 is a solution to (A). The
transform (x-g)/2 now gives a prime p.
Which checks out o.k.
Estimate of number of [a,b,c] with c<=N.
Admissible mod 6: [0,0,0],[2,2,2],[4,4,4],[2,0,0],[0,2,0],[0,0,2],[4,0,0],[0,4,0],[0,0,4],[2,2,0],[2,0,2],[0,2,2],[4,4,0],[4,0,4]
There are 15 combinations [a,b,c] with are admissible mod 6 (or 3),
out of 3^3=27 combinations.
About half of the sums a+b+c are 2 mod 4.
One can choose N/2 different even numbers 0<n<N.
The total number of [a,b,c] to choose from is smaller than
(N/2)^3/6=N^3/48. [Can be made precise].
So I would estimate this with N^3/48*15/27*1/2 = 5/864*N^3.
Some have to be descarded because we need d to have a square root of
2. Testing reveals that although we can discard about half the d,
this is not true for the [a,b,c], the reason is that we have 3
d-values (depending on the type of equation) for 1 set of [a,b,c].
P(keep [a,b,c] for any type)=1-P(discard [a,b,c] for one type)^3=7/8
Leading to the final estimate 35/6912*N^3, which is too large, but
this gives a fair clue to the amount of work needed.
All d>0 are of type I. So type II,II have d<0.
The situation d=-1 is possible. In fact I found 138 different values
of g belonging to this d. All are type I.
It would be helpful if one could prove that all these periodic
continued fractions have period 1 with periodic part 2.
Because d and g comprise 2 parameters derived from 3 parameters [a,b,c]
this routine actually checks more combinations [a,b,c] than those
used to compute d and g.
Either broaden the horizon and allow a larger set of (d,g) or
increase the number of iterates (to check larger prime solutions) or
Better would be to try and find a deeper relation to restrict the
As a sidenote I used Maple. But itís memory management went Ďout of
controlí. So taking a next step would entail rewriting the routine
in c, giving me more control on this aspect.
I changed my routine:
- The admissible d are computed
differently. d is admissible if and only if d has prime divisors
p=+/-1 mod 8.
- It only computes roots r with
0<=r<=d/2 and uses these to find the primitive solution by using
the associated continued fraction of (r+sqrt(2))/|d|.
- This primitive solution is used
to find solutions for roots r in -d/2<=r<=d/2, upon using the
fundamental solution of the Pell equation x^2-2y^2=1.
A primitive solution,
x0+y0*sqrt(2) is multiplied by eta^n, with eta=3+2*sqrt(2), the
Or, for negative r,
x0-y0*sqrt(2) or -x0+y0*sqrt(2) is used, depending on the sign
- One iteration using the new
method has the same effect as 2 iterations using the old method
(and is faster because one can precompute the values of eta^n).
I found the following solutions
with a<b<c<1000 with a maximum of 200 iterations:
Type I) p*q+r*s=m^2:
Type II) no solutions, given the
search area and maximum # of iterations.
Type III) p*s+q*r=m^2:
I checked all primes with PRIMO.