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Thread: Show that x^2 is congruent to 1 (mod 24) given x congruent to 5 (mod 6)

  1. #1
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    Show that x^2 is congruent to 1 (mod 24) given x congruent to 5 (mod 6)

    Hi all,
    I will write out the word congruent instead of the symbol, as we use three lines over one another and I could not find a way to write it out.

    The question: "Let x be congruent to 5 (mod 6). Show that x^2 is congruent to 1 (mod 24)."

    I believe I need to set up one equation as indices, and somehow set them equal to each other?
    Such as if x is congruent to 5 (mod 6), one could write that as x = 6k +5 and x^2 = 24j + 1.

    Any help would be greatly appreciated!
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  2. #2
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    Re: Show that x^2 is congruent to 1 (mod 24) given x congruent to 5 (mod 6)

    If x= 6k+ 5 then x^2= 36k^2+ 60k+ 25= (36k^2+ 60k+ 24)+ 1= 12(3k^3+ 5k+ 2)+ 1. Now you need to show that 3k^2+ 5k+ 2 is even for all k. If k is even, that's easy- if k= 2j then 3k^2+ 5k+ 2= 12j^2+ 10j+ 2= 2(6j^2+ 5j+ 1). If k is odd, k= 2j+ 1 and then

    3(4j^2+ 4j+ 1)+ 5(2j+ 1)+ 2= 12j^2+ 12j+ 3+ 10j+ 5+ 2= 12j^2+ 22j+ 10= 2(6j^2+ 11j+ 5).
    Thanks from Plato, topsquark and azollner95
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