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Math Help - find all prime numbers

  1. #1
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    find all prime numbers

    find all prime numbers p,q,r such that (p/q )- (4/(r+1))=1
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  2. #2
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    Quote Originally Posted by nh149 View Post
    find all prime numbers p,q,r such that (p/q )- (4/(r+1))=1
    Hi!

    We first rewrite the equality \frac{p}{q}-\frac{4}{r+1}=1 equivalently as (p-q)(r+1)=4q.
    If r=2, we get 3p=7q, which forces q=3 and p=7, so we've found one triplet.
    So suppose from now on that r is an odd prime.
    If q=2 we have (p-2)(r+1)=8. We can't have r=3 because it implies p=4. Forced is r=7 and p=3, which gives us another triplet.
    So suppose from now on that q is an odd prime.
    From (p-q)(r+1)=4q we see that p is an odd prime.

    We now know 2|(p-q), 2|(r+1) so we can rewrite (p-q)(r+1)=4q as \frac{p-q}{2}\cdot \frac{r+1}{2}=q.

    We cannot have \frac{r+1}{2}=1 (else r wouldn't be a prime) so we must have \frac{r+1}{2}=q and \frac{p-q}{2}=1.

    In other words, p-q=2 and r=2q-1.


    Let q \equiv l \mod{3} for some l\in \{0,1,2\}. Then p \equiv l+2 \mod{3} and r \equiv 2l-1 \mod{3}.

    If l=0, we have q=3, p=5 and r=5, so we've found another triplet.
    If l=1, then p=3 which implies q=1. So we cannot have l=1.
    If l=2, then r=3 which implies q=2, which we excluded, so we cannot have l=2.

    We conclude that the only triplets are 7,3,2 and 3,2,7 and 5,3,5.
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