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Can you answer this conjecture? I have plenty of evidence to suggest it is correct.?
a, b, and k are positive integers. k is a prime number (not including 1). 0<a<k and 0<b<k. Prove that k is not a factor of ab.
Tomp gives an interesting answer, which I am studying. It will take me time to go through it.
However, one flaw might be that a or b could be prime numbers, though each less than k. I do not know if the proof assumes that a and or b are composite.
Perhaps, if that is the case, the proof may not work.
2 Answers
- TompLv 71 decade agoFavorite Answer
Let the prime factorisation of a be
a = p1^m1 * p2^m2 * ...... * pr^mr
where pi is one of the prime factors of a raised to the power of mi and pi <= a < k
Similarly for b
b = q1^n1 * q2^n2 * ...... * qs^ns
where qj is a prime factor of b raised to the power of nj and qj <=b < k
Note that the above two factorisations are unique to a and b.
Now
ab = (p1^m1 * p2^m2 * ...... * pr^mr)(q1^n1 * q2^n2 * ...... * qs^ns)
Since k is prime, none of the prime factors of a or b can divide k.
Conversely k cannot divide pi (1 <= i <= r) nor qj (1 <= j <= s)
Hence k cannot be a factor of the product ab
By the way, 1 is not a prime number. A prime number k is defined by the fact that it can only have two divisors, 1 and k itself.. The number 1 has only one divisor, 1 itself.
- lienad14Lv 61 decade ago
the only POSITIVE PRIME is 2 Therefore k = 2.
If a and b are positive and less than 2 then they do not exist
