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Thread: Number theory problem proof

  1. #1
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    Number theory problem proof

    The problem is: Prove that 7|x2+y2only if 7|x and 7|y for x,yZ
    I found a theorem in my book that allows to do the following transformation: if a|band a|c-> a|(b+c)

    So, can I prove it like this: 7|x2+y2=>7|x2,7|y2=>7|xx,7|yy=>7|x,7|y?
    I am not really sure because I have this simple example in my head that even if 6|18 => 6|14+4 -> 614 and 64.
    Any help will be appreciated, thank you!
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  2. #2
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    Re: Number theory problem proof

    Quote Originally Posted by Breh View Post
    The problem is: Prove that 7|x2+y2only if 7|x and 7|y for x,y∈Z
    I found a theorem in my book that allows to do the following transformation: if a|band a|c-> a|(b+c)
    So, can I prove it like this: 7|x2+y2=>7|x2,7|y2=>7|x∗x,7|y∗y=>7|x,7|y?
    I am not really sure because I have this simple example in my head that even if 6|18 => 6|14+4 -> 6∤14 and 6∤4.
    When posting please do not use fancy fonts and colors. I just remove them as I did above.

    A lesson in logic. The statement that $P\Rightarrow Q$ is equivalent to $P$ only if $Q$.
    Then in this question we assume that $7|(x^2+y^2)$ and try to show that $7|x~
    \& ~7|y$.

    Now show some of your work.
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  3. #3
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    Re: Number theory problem proof

    Sorry for not being fancy enough. "Then in this question we assume that $7|(x^2+y^2)$ and try to show that $7|x$ & $7|y.$" Dude, that's the problem. Thank you for retyping, I appreciate the help.
    Last edited by Breh; Jun 4th 2017 at 12:37 PM.
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  4. #4
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    Re: Number theory problem proof

    Since you spend a lot of time online "helping" people, the answer is on page 5 in http://math.uga.edu/~pete/4400twosquares.pdf . Read it and give an actual help to the next person who asks for it.

    All the best.
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  5. #5
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    Re: Number theory problem proof

    Quote Originally Posted by Breh View Post
    Sorry for not being fancy enough. "Then in this question we assume that $7|(x^2+y^2)$ and try to show that $7|x$ & $7|y.$" Dude, that's the problem. Thank you for retyping, I appreciate the help.
    Quote Originally Posted by Breh View Post
    Since you spend a lot of time online "helping" people, the answer is on page 5 in http://math.uga.edu/~pete/4400twosquares.pdf . Read it and give an actual help to the next person who asks for it.
    All the best.
    My dear man/women what did you read into my reasonably polite reply that cause such venom on your part? It is after all a requirement of this forum that one asking for help must show some effort. We are not simply a homework service.
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  6. #6
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    Re: Number theory problem proof

    I agree. I didn't ask for the solution but for an advice if I am doing it right, because I started something and I even asked if that is correct, didn't I? Your reply wasn't polite at all, you are arrogant and not very helpful
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  7. #7
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    Re: Number theory problem proof

    Quote Originally Posted by Breh View Post
    I agree. I didn't ask for the solution but for an advice if I am doing it right, because I started something and I even asked if that is correct, didn't I? Your reply wasn't polite at all, you are arrogant and not very helpful
    Wow, Plato! (S)he's got your number! How dare you ask him/her for clarification of his/her question, or to format his/her question to be easily read? The audacity!

    Breh, I'm sure you feel that in your 16 comments on this forum, you have developed a mastery of helping on Math Help Forums. I assure you that is not the case. Plato has helped with literally thousands of posts. By what authority could you possibly tell him that his methods of helping are unsatisfactory? Seriously, if this is how you behave I certainly would not want to offer my help.
    Last edited by SlipEternal; Jun 4th 2017 at 07:00 PM.
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  8. #8
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    Re: Number theory problem proof

    A solution was posted so I consider the question to be answered.

    I would like to say that Plato was doing a good job and that I hope he will continue to do the work in the way that he's been doing it all along.

    -Dan
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