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Is there a procedure to solve this or is it strictly by trial and error?

$5^x \equiv 5^y \pmod {39}$ where $y > x$.

Thanks.

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    What do you mean by "solved"? There are infinite answers for both x and y with modular arithmetic.2012-02-14
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    Essentially, the first element to repeat in the sequence: 5^1 mod 39, 5^2 mod 39, 5^3 mod 39.....2012-02-14
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    Answer: `x = 0, y = 4` EDIT: `x = 4, y = 8`2012-02-14
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    x and y are greater than or equal to one2012-02-14
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    IIRC, these are called/related to [Discrete Logarithms](http://en.wikipedia.org/wiki/Discrete_logarithm). Which aren't easy compute.2012-02-14
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    @Mysticial: This problem is much easier than the general discrete logarithm problem. Not only because the numbers are small, but mostly because here we want $x$ in terms of $y$. See the answers below.2012-02-14

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