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Math Help - Modulo of squares = modulo of roots

  1. #1
    Member oldguynewstudent's Avatar
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    Modulo of squares = modulo of roots

    Stuck again! By the way thanks again for all the help!

    Prove or disprove:

    For all positive integers a, b: if a^2 \equiv b^2 (mod 9), then a \equiv b (mod 9)

    I haven't found a counterexample so it looks true and I have the proof for the converse. I've only gotten this far:

    Given
    a^2 \equiv b^2 (mod 9)

    then 9 | ( a^2 - b^2) = (a - b)(a + b)

    I know that if s | tu then s | t or s | u.
    So 9 | (a - b) or 9 | (a + b) and we need to show that 9 | (a - b) but I'm not sure where to go from here.

    Thanks
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    Stuck again! By the way thanks again for all the help!

    Prove or disprove:

    For all positive integers a, b: if a^2 \equiv b^2 (mod 9), then a \equiv b (mod 9)

    I haven't found a counterexample so it looks true and I have the proof for the converse. I've only gotten this far:

    Given
    a^2 \equiv b^2 (mod 9)

    then 9 | ( a^2 - b^2) = (a - b)(a + b)

    I know that if s | tu then s | t or s | u.
    So 9 | (a - b) or 9 | (a + b) and we need to show that 9 | (a - b) but I'm not sure where to go from here.

    Thanks
    Surely 1 and, for example, 8 work...
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  3. #3
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    Quote Originally Posted by oldguynewstudent View Post
    Stuck again! By the way thanks again for all the help!

    Prove or disprove:

    For all positive integers a, b: if a^2 \equiv b^2 (mod 9), then a \equiv b (mod 9)

    I haven't found a counterexample so it looks true and I have the proof for the converse. I've only gotten this far:

    Given
    a^2 \equiv b^2 (mod 9)

    then 9 | ( a^2 - b^2) = (a - b)(a + b)

    I know that if s | tu then s | t or s | u.
    So 9 | (a - b) or 9 | (a + b) and we need to show that 9 | (a - b) but I'm not sure where to go from here.

    Thanks
    "I know that if s | tu then s | t or s | u." - This is wrong - and rather a grave mistake. I will let you think about it.
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  4. #4
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    Grandad's Avatar
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    Hello oldguynewstudent

    Following the previous replies, you might also like to work out why the following are also counterexamples:
    a=2, b= 7

    a=3,b=6

    a=4, b=5
    Grandad
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  5. #5
    Member oldguynewstudent's Avatar
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    Yes, I see that now

    Quote Originally Posted by aman_cc View Post
    "I know that if s | tu then s | t or s | u." - This is wrong - and rather a grave mistake. I will let you think about it.
    Yes, it's the converse that's true, if a|b then a | bc
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