Prove that if a is an odd integer, then a^2 is congruent to 1(mod8)
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Originally Posted by mandy123 Prove that if a is an odd integer, then a^2 is congruent to 1(mod8) $\displaystyle a=8k\pm 1,8k\pm 3$. Now square them.
so when I square them how exactly is that going to help me prove that a is congruent to 1(mod8)? Am I just suppose to give examples like if k=3 then it would be 625 which is congruent to 1(mod8) and so on, or do I need to continue the proof?
Originally Posted by mandy123 so when I square them how exactly is that going to help me prove that a is congruent to 1(mod8)? $\displaystyle (8k\pm 1)^2 = 8(8k^2) + 8(\pm 2k) + 1 = 8(8k^2 \pm 2k)+1$ thus it is congruent to 1 mod 8. $\displaystyle (8k\pm 3)^2 = 8(8k^2) + 8(\pm 6k) + 9 = 8(8k^2 \pm 6k + 1)+1$ thus it is congruent to 1 mod 8.
oh, ok, i see now how you did it. I was just squaring them completely differently than you. Thank you so much!
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