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STEP Maths Question Watch

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    Please refer to Advanced Problems in Core Mathematics by Stephen Siklos, page 24 (question 24, part iii) here.

    Now that we know n and a are both even, we can follow the method used in part (ii) and set n = 2m and a = 2b . This gives
    (2m − 2b)^3 + (2m)^3 = (2m + 2b)^3
    from which a factor of 2^3 can be cancelled from each term. Thus m and b satisfy the same equation as n and a. They are therefore both even and we can repeat the process.
    Repeating the process again and again will eventually result in an integer that is odd which will therefore not satisfy the equation that it is supposed to satisfy: a contradiction. There is therefore no integer n that satisfies the equation
    I repeated the process once by taking out the factor 2^3 from the equation, giving (m-b)^3 + m^3 = (m+b)^3, where m and b are both even/odd. So how am I supposed so obtain "integer that is odd which will therefore not satisfy the equation"? If both m and b are even, it will result in a repetition of part ii, which consequently results in part i, while b being odd will result in part i right away.

    So... how do I justify the bold statement above? Thank you!
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    If you keep dividing by two you eventually get an odd number.
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    Considering m in terms of prime factors should help you see why it works.
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    (Original post by Zuzuzu)
    Considering m in terms of prime factors should help you see why it works.
    Could you please elaborate? Thank you.
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    (Original post by johnconnor92)
    Could you please elaborate? Thank you.
    m = 2^n \cdot \text{(at least one odd prime)}

    Repeated division by 2 should give you an odd number eventually.
 
 
 
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