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    I guess one way of doing it would be:

    even number = 2x
    next even number = 2x+2

    sum of squares:

    (2x)²+(2+2x)²
    4x²+4x²+8x+4
    8x²+8x+4
    4(2x²+2x+1)

    to be divisible by 8, (2x²+2x+1) must be divisble by 2

    2(x²+x)+1 which is going to be odd whatever the value of x... and therefore does NOT go into 2... therefore the sum of the squares of any two consecutive even integers is never a multiple of 8 and always a multiple of 4 :P
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    (Original post by Feria)
    I guess one way of doing it would be:

    even number = 2x
    next even number = 2x+2

    sum of squares:

    (2x)²+(2+2x)²
    4x²+4x²+8x+4
    8x²+8x+4
    4(2x²+2x+1)

    to be divisible by 8, (2x²+2x+1) must be divisble by 2

    2(x²+x)+1 which is going to be odd whatever the value of x... and therefore does NOT go into 2... therefore the sum of the squares of any two consecutive even integers is never a multiple of 8 and always a multiple of 4 :P
    This guy posted the same thing 3 times. it has already been answered feria. the easier way is to go from your step of:
    8x²+8x+4 = 8(x²+x) +4. cannot be multiple of 8. although your method is still completely correct.
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    let the two consectutive even integers be 2k and 2k+2.

    S = (2k)² + (2k+2)²
    S = 4k² + 4k² + 8k + 4
    S = 8k² + 8k + 4
    S = 8(k² + 1) + 4

    division of S by 8 will always leave a remainder of 4

    ergo, S is never a multiple of 8
    ========================
 
 
 
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