JM Bergot sent
the following observation:
78*79=6162 in which one has on the left a product
of p and p-1 and on the right the concatenation of prime q and
Q. Can you
report similar larger cases & near variations?
Contributions came from Torbjörn Alm, Hakan Summakoglu, Robert D Mohr,
J-C Rosa &
Emmanuel Vantieghem (the best solution!!!).
Torbjörn Alm wrote:
The only two solutions below p<10^9, the first 60000000 primes are
79*78 = 6162
5232869*5232868 = 27382912738292
Number of solutions: 2
5*7=35 (p,p+2 and q,q+2 are twin primes)
Here are all the solutions I found, including the trivial ones
and those where p is composite.
I only found one example where p was prime -- p=5232869:
3 * 4 = 12
7 * 8 = 56
78 * 79 = 6162
80919 * 80920 = 6547965480
91809 * 91810 = 8428984290
326510 * 326511 = 106609106610
475025 * 475026 = 225649225650
524975 * 524976 = 275599275600
673490 * 673491 = 453589453590
4323777 * 4323778 = 18695051869506
4767132 * 4767133 = 22725552272556
5232868 * 5232869 = 27382912738292
<------- This is the only one where p is prime
5676223 * 5676224 = 32219513221952
4083911141 * 4083911142 = 16678330211667833022
<------ Here p-1 is prime
4975000250 * 4975000251 = 24750627492475062750
5024999750 * 5024999751 = 25250622492525062250
5916088859 * 5916088860 = 35000107393500010740
9503960496 * 9503960497 = 90325265119032526512
9604950396 * 9604950397 = 92255072119225507212
It seems odd to me that there are four 7-digit solutions and 6
10-digit solutions but none for 8- or 9-digits.
I also find it curious that they seem to show up in pairs.
1) for p(p-1)=q&(q+1) ( p and q primes)
p=79 ; q=61 (already published)
p=5232869 ; q=2738291
2) for p(p+1)=q&(q+1) I found only two solutions:
p=7 ; q=5 ( very nice because with p+1 and q+1 we have four consecutive
p=4083911141 ; q=1667833021
There are only three solutions (p,q) of p(p-1) =
q&(q+1) with p fewer than 72 digits :
(79, 61), (5232869, 2738291) and
The last p has 70 digits.
Finding bigger solutions requires factorizations of numbers with
more than 73 digits and the scarceness of the solutions makes it a bit
uncertain that a bigger solution is within reach.
The only 'near variations' I considered are :
p(p+1) = q&(q+1) with only one solution (p,q) with less than
72 digits : ( 4083911141, 1667833021 ).
The variations p(p-1) = q&(q-1) and p(p+1) = q&(q-1) cannot
The first problem will have solutions iff the quadratic
polynomial x^2 - x - 1 has a root modulo 10^n+1, where n is the
number of digits of p.
Assume there are integers p, q (not necessary prime) such that
p(p-1) = q&(q+1). Then we can write p(p-1) = q10^n+q+1, where n is
the number of digits of p. But then, we can write p(p-1)-1 =
q(10^n+1), whence the polynomial x(x-1) - 1 = x^2-x-1 has a zero
Since the discriminant of the polynomial is 5, we only have to
look for those n such that 5 is quadratic residue
modulo 10^n+1. With Mathematica, the search for the solutions goes very
quick. Only a filtering for primes solutions completes the job.
For the last two 'near variations', the discriminant is -3 and
that is never quadratic residue modulo a number of the form 10^n+1 (of
course, this argumentation requires the quadratic reciprocity law for
the Jacobi symbol).