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Math Help - Euler's Criterion

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    Euler's Criterion

    I am trying to show that x^2 == 25(mod 997) has a solution, but I can't seem to get there.
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    Member Black's Avatar
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    x=5?
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    Yeah, thanks. Caught onto that after I belatedly realized that 997 is prime. Thank you!
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    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by tarheelborn View Post
    Yeah, thanks. Caught onto that after I belatedly realized that 997 is prime. Thank you!
    It doesn't matter in this case that 997 is prime; you'd still have the solution x=\pm 5 (although perhaps you'd have others as well).
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    But since 997 is prime, wouldn't Lagrange's theorem apply to show that there are ONLY 2 solutions?
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    Quote Originally Posted by tarheelborn View Post
    But since 997 is prime, wouldn't Lagrange's theorem apply to show that there are ONLY 2 solutions?


    What has Lagrange to do here? Since 997 is a prime there are only two solutions because any polynomial p(x) ( e.g., x^2-25=0 ) over a field can have at most \deg(p(x)) different roots in some extension of the field.

    Tonio
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    Quote Originally Posted by Bruno J. View Post
    It doesn't matter in this case that 997 is prime; you'd still have the solution x=\pm 5 (although perhaps you'd have others as well).
    Sure, I see that; I was just trying to clarify your statement that perhaps you would have other solutions as well. Thank you so much for your help!
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