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Math Help - Perfect square

  1. #1
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    Arrow Perfect square

    1) If gcd(x,3)=1 and gcd(y,3)=1, prove that x^2 + y^2 cannot be a perfect square.

    2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.

    This is not homework, I'm doing it for extra practice, but I can't figure out how to prove these.

    Any help would be appreciated!

    [also under discussion in math link forum]
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    1) If gcd(x,3)=1 and gcd(y,3)=1, prove that x^2 + y^2 cannot be a perfect square.
     (x,3)=1\implies x\equiv\pm1\bmod{3}

    Therefore  x^2+y^2\equiv1+1=2\bmod{3} .

    So if  x^2+y^2=z^2 \implies z^2\equiv 2\bmod{3} , which is impossible.
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  3. #3
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    Quote Originally Posted by kingwinner View Post
    2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.
    Fix a prime p, suppose the exponent of a,b,c of p are x,y,z respectively, WLOG suppose a<=b<=c. Show a=b=1
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  4. #4
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    Quote Originally Posted by FancyMouse View Post
    Fix a prime p, suppose the exponent of a,b,c of p are x,y,z respectively, WLOG suppose a<=b<=c. Show a=b=1
    Q2) I've thought about it for awhile, but I can't figure out how to show a=b=1.

    Also, aren't we supposed to show gcd(a,b)=gcd(b,c)=gcd(a,c)=1. How can we prove this?

    Thanks!
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  5. #5
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    Quote Originally Posted by chiph588@ View Post
     (x,3)=1\implies x\equiv\pm1\bmod{3}

    Therefore  x^2+y^2\equiv1+1=2\bmod{3} .

    So if  x^2+y^2=z^2 \implies z^2\equiv 2\bmod{3} , which is impossible.
    Thanks!
    How about this?
    "If x and y are odd, prove that x^2 + y^2 cannot be a perfect square."

    x odd => x=1(mod2) => x^2 = 1 (mod 2)
    => x^2 + y^2 = 0 (mod 2)
    => ???
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  6. #6
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    Thanks!
    How about this?
    "If x and y are odd, prove that x^2 + y^2 cannot be a perfect square."

    x odd => x=1(mod2) => x^2 = 1 (mod 2)
    => x^2 + y^2 = 0 (mod 2)
    => ???
     (x,3)=1\not\Longrightarrow x is odd. For example,  (4,3)=1 .
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  7. #7
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    Quote Originally Posted by chiph588@ View Post
     (x,3)=1\not\Longrightarrow x is odd. For example,  (4,3)=1 .
    Sorry, maybe I wasn't clear. This is a different problem.
    "If x and y are odd, prove that x^2 + y^2 cannot be a perfect square."
    How can we prove this?
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    Sorry, maybe I wasn't clear. This is a different problem.
    "If x and y are odd, prove that x^2 + y^2 cannot be a perfect square."
    How can we prove this?
    Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of  2 .
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  9. #9
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    Quote Originally Posted by chiph588@ View Post
    Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of  2 .
    ""If x and y are odd, prove that cannot be a perfect square."

    How can we prove this directly by basic modular arithmetic?

    x odd => x=1(mod2) => x^2 = 1 (mod 2)
    => x^2 + y^2 = 0 (mod 2)
    But I can't find a contradiction here...
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  10. #10
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kingwinner View Post
    ""If x and y are odd, prove that cannot be a perfect square."

    How can we prove this directly by basic modular arithmetic?

    x odd => x=1(mod2) => x^2 = 1 (mod 2)
    => x^2 + y^2 = 0 (mod 2)
    But I can't find a contradiction here...
    This isn't a contradiction since  z^2\equiv0\bmod{2} is solvable.
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  11. #11
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    Quote Originally Posted by chiph588@ View Post
    Look at the definition of a Pythagorean triple (not necessarily primitive). One term is always a multiple of  2 .
    So how can we prove this claim?
    (this question is from the section on basic modular arithmetic; Pythagroean triple hasn't been introduced yet so I don't think we'll need it)

    Thanks!
    Last edited by kingwinner; April 9th 2010 at 08:00 PM.
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  12. #12
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    So how can we prove this claim?
    (this question is from the section on basic modular arithmetic; Pythagroean triple hasn't been introduced yet so I don't think we'll need it)

    Thanks!
    Why can't you just note that if x=2n+1,y=2m+1 then x^2+y^2=4n^2+4n+1+4m^2+4m+1\equiv2\text{ mod }4 but the reduced residue class of x^2\text{ mod }4 is \{0,1\}?
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  13. #13
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    Thanks! How to prove Q2?
    2) Given that gcd(a,b,c)lcm(a,b,c) = abc, prove that gcd(a,b)=gcd(b,c)=gcd(a,c)=1.
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  14. #14
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    Quote Originally Posted by kingwinner View Post
    Q2) I've thought about it for awhile, but I can't figure out how to show a=b=1.

    Also, aren't we supposed to show gcd(a,b)=gcd(b,c)=gcd(a,c)=1. How can we prove this?

    Thanks!
    Sorry there's a typo. I meant x=y=1, not a=b=1. i.e. if p^x || a (i.e. p^x divides a but p^{x+1} doesn't divide a), p^y || b, p^z || c, and x<=y<=z, then x=y=1, which implies that if some prime one of a,b,c is divisible by p, then only one of them is divisible by p
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