modular arithmetic with exponents

Question

I'm looking at the solution manual of a book and it lists a solution for

$$[19^3\mod {23}]^2 \pmod {31} \equiv [(-4)^3\mod {23}]^2 \pmod {31} \equiv [-64\mod {23}]^2 \pmod {31}\equiv 5^2 \pmod{31} = 25$$

How does it get from $[19^3\mod {23}]^2 \pmod {31}$ to $[(-4)^3\mod {23}]^2 \pmod {31}$?

in other words where does the transformation from $19^3$ to $(-4)^3$ come from?

$19 \pmod {23} = -4 \pmod {23} $ – Amzoti Nov 09 '13 at 05:33 — Amzoti, Nov 09 '13 at 05:33

score 1 · Accepted Answer · answered Nov 09 '13 at 05:35

1

The important thing is that it is from $19^3 \pmod {23}$ to $(-4)^3 \pmod {23}$ This works because $19 \equiv -4 \pmod {23}$ You lost that in going to the final question.

answered Nov 09 '13 at 05:35

Ross Millikan

374,822

Ahh. All the exponents were distracting me from the concept the problem was trying to stress (intentionally I suppose). Thanks! – user609926 Nov 09 '13 at 05:51

modular arithmetic with exponents

1 Answers1