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    I'm stuck on my reasoning for this algebraic proof:
    Prove algebraically that the sum of the squares of any two consecutive numbers always leaves a remainder of 1 when divided by 4.

    My working:

    Two consecutive numbers = n and n+1
    (n)^2 + (n+1)^2 = n^2+n^2+n+n+1
    = 2n^2+2n+1
    I've then factorised the 2n out to get:
    2n(n+1)+1

    How do I then explain my reasoning of why there is a remainder of 1 when the term is divided by 4?
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    (Original post by Nikey)
    I'm stuck on my reasoning for this algebraic proof:
    Prove algebraically that the sum of the squares of any two consecutive numbers always leaves a remainder of 1 when divided by 4.

    My working:

    Two consecutive numbers = n and n+1
    (n)^2 + (n+1)^2 = n^2+n^2+n+n+1
    = 2n^2+2n+1
    I've then factorised the 2n out to get:
    2n(n+1)+1

    How do I then explain my reasoning of why there is a remainder of 1 when the term is divided by 4?
    If you can show that n(n+1) is even then that must mean that 2n(n+1) is a multiple of 4. Think about why n(n+1) must be even.
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    (Original post by Notnek)
    If you can show that n(n+1) is even then that must mean that 2n(n+1) is a multiple of 4. Think about why n(n+1) must be even.
    Because either n or n+1 must be even, so the other must be odd. Even number × odd number = even number. Is that it?
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    (Original post by Nikey)
    Because either n or n+1 must be even, so the other must be odd. Even number × odd number = even number. Is that it?
    Yes that's right.
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    (Original post by Notnek)
    Yes that's right.
    Okay, thanks for helping
 
 
 
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