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

1. 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!

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

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

$\displaystyle 3(4j^2+ 4j+ 1)+ 5(2j+ 1)+ 2= 12j^2+ 12j+ 3+ 10j+ 5+ 2= 12j^2+ 22j+ 10= 2(6j^2+ 11j+ 5)$.