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p and q are odd primes such that p>q. Does the equation (k-p)q= k-1 have any natural number k as the solution? I have proved that no natural number up to p+1 satisfies the equation. I am unable to come to a conclusion regarding natural numbers beyond p+1. Kindly help me solve this problem. Thanks.

tony
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  • $k=19$, $p=13$ $q=3$ – lulu Aug 05 '16 at 10:42
  • I came across this equation trying to solve a group theory problem.Can you help me with it?http://math.stackexchange.com/questions/469871/existence-of-group-of-order-p-in-group-of-order-pq-pq?s=14|2.3091 – tony Aug 05 '16 at 10:51
  • Looking at the argument in that link, I don't see any elementary way to complete it. Maybe I am missing something. You could write to that author and ask. Cauchy's Theorem really seems like the way to go with that exercise. I can probably invent an argument specific to groups of order $pq$ but I doubt it will be significantly easier than the general case. – lulu Aug 05 '16 at 11:12
  • Thanks. I would have asked but I lack points to comment on other people's questions. – tony Aug 05 '16 at 11:25
  • here is another problem which claims to address groups of order $pq$ separately. If you look through the answers to that I think you'll find leads on how to do it without the general theory. I have not gone through them carefully myself, however, so can't attest to their accuracy or completeness. – lulu Aug 05 '16 at 11:33

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At least one solution exists $$k=7,q=3,p=5$$ $$(7-5)3=7-1$$ $$6=6$$

Wouter
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  • Thanks. I actually came to this equation trying to solve an algebra problem. Can you help me with it. http://math.stackexchange.com/questions/469871/existence-of-group-of-order-p-in-group-of-order-pq-pq?s=14|2.3091 – tony Aug 05 '16 at 10:46