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Math Help - Proofs-squares

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    Proofs-squares

    prove that the sum of two odd squares cannot be a square
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    Quote Originally Posted by anncar View Post
    prove that the sum of two odd squares cannot be a square
    Let the two odd numbers be x = 2m + 1 and y = 2n + 1

    Sum the two squares and note that the result is divisible only by 2, not 4.

    Now let any even number be z = 2p. The square of this number is divisible by 4.

    etc, etc.

    -Dan
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    Hello, anncar!

    A variation of Dan's explanation . . .


    Prove that the sum of two odd squares cannot be a square.

    If n is even, n = 2k, its square is: . n^2 \:=\:(2k)^2\:=\:4k^2
    . . The square of an even number is a multiple of 4. .[1]

    If n is odd, n = 2k+1, its square is: . n^2\:=\:(2k+1)^2 \:=\:4(k^2 + k) + 1
    . . The square of an odd number is one more than a multiple of 4. .[2]

    These are the only two forms for the square of an integer.


    Let the odd integers be: . a \:=\:2h+1,\;b\:=\:2k+1

    The sum of their squares is: . N \;=\;a^2+b^2 \;=\;(2h+1)^2+(2k+1)^2

    . . =\:4h^2+4h + 1 + 4k^2 + 4k + 1 \:=\:4(h^2+h+k^2+k)+2

    Hence, N is two more than a multiple of 4.


    Since N cannot be either of the forms [1] or [2], N cannot be a square.

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