Problems & Puzzles: Puzzles

Puzzle 570.- Does this happen again?

JM Bergot sent the following puzzle:

For three consecutive primes p<q<r, one has q*r=Amod(p), p*r=Bmod(q), and p*q=Cmod(r).  For 11,13,17,  A+B+C =13, which is one of the three primes.

Q.  Does this ever happen again?

Contributions came from Torbjörn Alm, Emmanuel Vantieghem & Farideh Firoozbakht

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Torbjörn Alm

Only other solution found: Solution for 2 3 5, A= 1 B=1 C =1 sum=3

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Emmanuel Vantieghem wrote:

Here is my contribution to Puzzle 570 :
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I think that  A+B+C = one of the primes  p, q, r  will not happen again.
If we put  q - p = v  and  r - q = w, then I found that, for  p > 1327,  A = v(v+w), C = w(v+w)  and  B = q-vw.
That means that in these cases, A+B+C >= r+24.  The value  24  occurs when  v = 2  and  w = 4.
Eilas, the conjecture that  v(v+w) < p  and  w(v+w) < r  for  p  big enough is a very strong one (stronger that Andrica's for instance), whence a proof will be far away.
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Sincere greeting and : happy  (123 - 4)*(5+6) + 78*9 (or  (9*8 + 7)*(6+4*5)-43*(2 - 1)  or  3^4+3^4+43*43) !

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Farideh Firoozbakht wrote:

For the three consecutive  primes 2, 3, 5 we have :

2*3 = 1 (mod 5), 3*5 = 1 (mod 2) , 5*2 = 1 (mod 3)   & 1 + 1 + 1 = 3.

I didn't find third solution yet. My search for finding it will be finished soon. If I find it
I will report  you.

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Andreas Höglund wrote:

I found no other occurence below 15*10^9 and I don’t think there is any more.
If we call the primes: p, p+a, p+b with b>a:
working (mod p+b):   p*(p+a) = p^2+a*p = (-b)^2+a*p = b^2+a*(-b) = b^2-a*b (mod p+b)
working (mod p+a):   p*(p+b) = p^2+b*p = (-a)^2+b*p = a^2+b*(-a) = a^2-a*b (mod p+a)
working (mod p):       (p+a)(p+b) = p^2+(a+b)p+a*b = a*b (mod p)
So the sum is: b^2-a*b+a^2-a*b+a*b = (b-a)^2 + a*b

Here a is the prime gap, and b is the sum of 2 consecutive prime gaps. As p gets larger prime gaps gets smaller compared to p, the largest gap is at most = 35*ln(p) where ln is the natural logarithm, http://users.cybercity.dk/~dsl522332/math/primegaps/gaps20.htm , so a*b will never again get to the size of p, p+a or p+b for p>11.

If we instead of consecutive primes allow the 3 primes to be within 5 primes of each other, that is within P(n) and P(n+4), there is a few more solutions: (p,q,r)=(3,7,13),(11,13,23),(17,19,29)
If we allow within 6 primes P(n) and P(n+5) there is also: (p,q,r)=(2,7,13),(3,5,17),(131,149,157)
so the further apart we allow them to be, the bigger solutions exists.

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