Prove that any integer of the form 3k+2 also has a prime factor of the form 3k+2............
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4 Answers
If it is a prime.. nothing to prove...
If it is not a prime.. and still even.. then 2 = 0.k+2 is a factor..
else if it is not even then k is odd.. hence k=2k'+1
so 3k+2=6k'+3+2
Now there is not much left to prove....
btw after writing this much there is a slightly simpler argument that i can think.
but can someone prove it from where i left?
Hint: if all prime divisors are of the form 3k+1, then the number is of the form?
simple one
after the hint in second post, it becomes an easy one
say a number `a` is of the form 3k+1
then a=1mod3
now if `a` does not contain a prime factor of the form 3k+2,
then all its prime factors should be of the form 3k+1 only
so number itself should beof the form 3l+1
that is ,
a=1mod3
but here , we have a is of the form 3k+2
means a=2mod3
so a contradiction ...
hence proved