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Math Help - on the subject of a+b=2c

  1. #1
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    on the subject of a+b=2c

    Hi again,

    I'm sure all the experienced users have come across this

    a^2+b^2=c^2

    let\ \ \ a=m^2-n^2

    and\ \ \ b=2mn

    then\ \ \ c=m^2+n^2

    so that you can choose any integers for m and n to get valid numbers for a, b & c.

    Has anyone come across anything similar for

    a^2+b^2=2c^2

    Thanks

    Pro
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: on the subject of a+b=2c

    I don't get your point. What's the intention of the substitutions a=m^2-n^2,...?
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  3. #3
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    Re: on the subject of a+b=2c

    Hi,

    Slightly off the point, but if you needed sides lengths for some pythagorean triangles, you could either trial and error your way with

    a^2+b^2=c^2

    or you could choose any m and n as given above to get valid results like this:

    let m=7 and n=9 (or any other integers you like)

    a=m^2-n^2=7^2-9^2=-32

    b=2mn=2*7*9=126

    c=m^2+n^2=7^2+9^2=130

    and you will see that

    32^2+126^2=130^2

    I am trying to find a similar way to break down

    a^2+b^2=2c^2
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  4. #4
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    Re: on the subject of a+b=2c

    Another method I just found on wolfram for obtaining Pythagorean triples using only 1 integer to start with by setting n to 1 is

    a=2m

    b=m^2-1

    c=m^2+1

    Has the same result since

    a^2 = 4m^2

    b^2 = m^4-2m^2+1

    c^2 = m^4+2m^2+1

    so\ \ \ a^2+b^2=c^2

    The trouble is

    a^2+b^2=2c^2

    is not Pythagoras
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  5. #5
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    Re: on the subject of a+b=2c

    Quote Originally Posted by procyon View Post
    Hi again,

    I'm sure all the experienced users have come across this

    a^2+b^2=c^2

    let\ \ \ a=m^2-n^2

    and\ \ \ b=2mn

    then\ \ \ c=m^2+n^2

    so that you can choose any integers for m and n to get valid numbers for a, b & c.

    Has anyone come across anything similar for

    a^2+b^2=2c^2

    Thanks

    Pro
    Let n,m be positive integers of opposite parity and n<m; also, let m and n be relatively prime.

    Then, all solutions of a^2+b^2=c^2 are precisely given by a=k(m^2-n^2), b=k(2mn), c=k(m^2+n^2), where k is any positive integer. Namely, any pair m,n satsifying the above restrictions and any k>0 will produce a Pythagorean Triple.

    Note that while what you wrote gives infinitely many solutions, not all solutions are produced this way (Example: the triple (15,36,39) will not show, and still 15^2+36^2=39^2).


    It happens that the question on (*) a^2+b^2=2c^2 has an answer depending on the above.

    From equation (*), we see that a,b are both odd or both even. So, the equation is equivalent to (\frac{a+b}{2})^2+(\frac{a-b}{2})^2=c^2 (simply verify this), where the fractions are actually integers.

    The problem has reduced to another problem that we already know to solve.
    There are two possibilities:

    (1) (\frac{a+b}{2})=k(m^2-n^2), (\frac{a-b}{2})=k(2mn), c=k(m^2+n^2). To find a and b explicitly we add and subtract (\frac{a+b}{2}) and (\frac{a-b}{2}) to get a=k(m^2-n^2+2mn), b=k(m^2-n^2-2mn).

    (2) (\frac{a+b}{2})=k(2mn), (\frac{a-b}{2})=k(m^2-n^2), c=k(m^2+n^2). In a similar way, a=k(m^2-n^2+2mn), b=k(n^2-m^2+2mn).

    Example, I choose m=5,n=2,k=1 and apply part(1): Then a=5^2-2^2+2\cdot5\cdot2=41,b=5^2-2^2-2\cdot5\cdot2=1,c=5^2+2^2=29. Indeed, 41^2+1^2=2(29^2).

    Or, if we applied part(2) with m=7,n=6,k=1: a=97,b=71,c=85 and it's true that 97^2+71^2=2(85^2).
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  6. #6
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    Re: on the subject of a+b=2c

    oops

    Excellent melese, it was so easy in the end.

    Many thanks for your help

    Pro
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