Odd Prime Triples: Find & Explore Solutions!

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SUMMARY

The discussion centers on finding all triples of odd primes (p, q, r) such that p^2 + 1 is divisible by q, q^2 + 1 is divisible by r, and r^2 + 1 is divisible by p. The known solutions include the triples (5, 13, 17) and (17, 29, 421). It is established that if p < q < r, there are no additional triples with p less than 10^7. The participants also note that all primes in these triples must be congruent to 1 modulo 4, as derived from quadratic residue theory.

PREREQUISITES
  • Understanding of prime numbers and their properties
  • Familiarity with modular arithmetic, specifically congruences
  • Knowledge of quadratic residue theory
  • Basic number theory concepts related to divisibility
NEXT STEPS
  • Research the properties of quadratic residues in number theory
  • Explore advanced topics in modular arithmetic
  • Investigate the implications of the BEAL Conjecture in relation to prime triples
  • Study computational methods for generating prime numbers and testing divisibility
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Mathematicians, number theorists, and students interested in prime number properties and divisibility rules will benefit from this discussion.

steiner1745
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I got this question from another
forum and it's driving me crazy.
Find all triples of odd primes,
p,q,r such that
p^2+1 is divisible by q, q^2+1 is divisible by r
and r^2+1 is divisible by p.
Two such triples are 5,13,17
and 17,29,421. If we assume
p<q<r, then there are no other
such triples with p<10^7.
Are there any others?
Anyone have any ideas?
From quadratic residue theory
we know that p,q,r are all
congruent to 1(mod 4).
Can we say more?
 
Last edited:
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(2,5,13) also works...
 
micromass said:
(2,5,13) also works...

nice observation but 2 is not an odd prime.

Years ago i thought i solved the BEAL CONJECTURE because i found 3^5 + 10^2 = 7^3

Then my math prof. pointed out ALL exponents must be integers greater than 2.:smile:
 

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