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17 April, 11:09

Show that if x, y, z are integers such that x^3+5y^3 = 25z^3, then x = y = 0.

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  1. 17 April, 11:20
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    3125*k^9 + y^3 is an integer my closure property.

    but 5^ (1/3) is not an integer, which forces z to be irrational.

    Note that there is no way an integer value can rationalize 5^ (1/3)

    Step-by-step explanation:

    x^3 = 25z^3 - 5y^3

    x^3 = 5 (5z^3 - y^3)

    x = (5 (5z^3 - y^3)) ^ (1/3) must be an integer

    = 5^ (1/3) * (5z^3 - y^3) ^ (1/3)

    Then (5z^3 - y^3) ^ (1/3) = 25*k^3 for some integer k

    5z^3 - y^3 = 15625*k^9

    5z^3 = 15625*k^9 + y^3

    z^3 = 3125*k^9 + (1/5) * y^3

    z = (3125*k^9 + (1/5) * y^3) ^ (1/3)
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