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 = qvw.
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^2a*b (mod p+b)
working (mod p+a): p*(p+b) = p^2+b*p = (a)^2+b*p =
a^2+b*(a) = a^2a*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^2a*b+a^2a*b+a*b = (ba)^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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