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Math Help - Algebraic Arguments

  1. #1
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    Exclamation Algebraic Arguments

    Hi Guys,
    I'm doing GCSE maths higher tier. And generally I'm quite good at algebra, subsitiution, factorising etc but I'd appreciate help with arguments.

    Here is the question:
    "Prove that the sum of squares of any two consecutive integers is an odd number".

    It has to be done algebraically. I have the answers but I don't understand them, can you please describe how you get the answer, also
    please can you provide general tips for answering these arguments mental approach, algebra etc. and the quicker the reply the better. I have a maths
    exam and it's quite urgent, so all help will be appreciated.

    Thanks,
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  2. #2
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    Re: Algebraic Arguments

    "Prove that the sum of squares of any two consecutive integers is an odd number".
    let x = initial integer

    x+1 = next consecutive integer

    x^2 + (x+1)^2 = x^2 + (x^2 + 2x + 1) = 2x^2 + 2x + 1 = 2(x^2 + x) + 1

    now, what can you say about the quantity 2(x^2+x) , where x is any integer?

    what happens when you add 1 to that quantity?
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  3. #3
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    Re: Algebraic Arguments

    Quote Originally Posted by curiousone View Post
    "Prove that the sum of squares of any two consecutive integers is an odd number".
    This is similar to proving that the sum of two consecutive integers is an odd number. This is, of course, because one of these integers is odd and the other is even. The reason the original question is similar is that the square of an even number is even and the square of an odd number is odd.

    Quote Originally Posted by curiousone View Post
    can you provide general tips for answering these arguments mental approach, algebra etc.
    When you are asked about two consecutive integers, you consider n and n + 1 for some n. An even number has the form 2k for some integer k, while an odd number has the form 2k + 1 for some integer k. For example, here is a proof that the square of an odd number is odd. Suppose you are given an arbitrary odd number. Then there must exist an integer k such that the given number is 2k + 1. Therefore, its square is (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2) + 1 = 2k' + 1 for k' = 2k2 + 2. Since k' is an integer, 2k' + 1 is odd.
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    Re: Algebraic Arguments

    Of what you said i understand x(squared)+(x{squared} +2x +1) from then on I'm lost.
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    Re: Algebraic Arguments

    Thanks for the help on the mental and algebraic bit
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    Re: Algebraic Arguments

    here's everything you need to know about even- and oddness:

    even + even = even
    even + odd = odd
    odd + odd = even

    (even)*(even) = even
    (even)*(odd) = even
    (odd)*(odd) = odd

    if we have two consecutive integers, one is even, and one is odd. therefore, their squares are (even)*(even) = even and (odd)*(odd) = odd.

    when we add their squares, we get: even + odd = odd.
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