Problems & Puzzles: Puzzles

Puzzle 592. p.(p-1)=q&(q+1) 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+1 Q. Can you report similar larger cases & near variations?

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Contributions came from Torbjörn Alm, Hakan Summakoglu, Robert D Mohr, J-C Rosa &  Emmanuel Vantieghem (the best solution!!!).

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Torbjörn Alm wrote:

The only two solutions below p<10^9, the first 60000000 primes are searched.

79*78 = 6162
5232869*5232868 = 27382912738292
Number of solutions: 2

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Hakan wrote:

5232868*5232869=27382912738292

Other variations: 
p.(p+2)=q&(q+2)

5*7=35 (p,p+2 and q,q+2 are twin primes)

p.(p+1)=q&(q+1)

7*8=56

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Robert wrote:

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.

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Rosa wrote:

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 numbers)
     p=4083911141  ; q=1667833021

Emmanuel wrote:

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

(3945871085890831895545081444727726758535879876730296127034028498958529, 1556989862646929286224773583956276832654440163733079609575209162885311).

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 have solutions! 

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 modulo 10^n+1.

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).

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