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I have maybe a simple question but I haven't been able to find an answer through googling.

The question is as such:

Is $\ a × b^2 \bmod c $ the same as $\ (a \bmod c) × (b^2 \bmod c) $?

That's it. That's all I'm confused about.

Sorry if it seems a little simple...

answer:

$\ a × b^2 \bmod c = ((a \bmod c) × (b^2 \bmod c))\bmod c $

:)

Bernard
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2 Answers2

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Where does the second product even take place? If it is over the integers then definitely no. $$2 \times 3^2 \pmod 6$$ is zero but $$2 \pmod 6$$ and $$3 \pmod 6$$ are just two and three. Thus their product will be 6.

If the second product is not taken over the integers, and everything is modulo c in the end, then it does not matter if you reduce $a$ and $b^2$ before or after you multiply.

Bernard
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user1390
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  • so essentially $\ 88 × 6^2 \ mod \ 25 $ is only the same as $( (88 \ mod \ 25) × (6^2 \ mod \ 25) ) \ mod \ 25$ aka - you modulo everything by $c $ in the end? –  May 26 '18 at 06:37
  • Correct! The rule that makes this work is often written as follows:

    If $a \equiv b$ and $c \equiv d$ then $ac \equiv bd$, where everything is modulo $n.$ This says that we can switch and multiply, or multiply and switch. So long as everything is modulo $n$ in the end you'll get the same result.

     – user1390 May 26 '18 at 14:07
  • thank you, that really clarifies it! –  May 26 '18 at 14:45
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Generally no. Distributive property here does not apply although there would be coincidental instance when it will hold.

Let

$ab^2\bmod c= r$

$a\bmod c= p$

$b^2\bmod c= q$

Then your question would imply that $r=pq$

Where $$r=\frac {ab^2}{lc}$$ $$p=\frac {a}{mc}$$ $$q=\frac {b^2}{nc}$$

$l,m,n$ are integers

Thus implies

$$r=\frac{ab^2}{lc}=(\frac{a}{mc})(\frac{b^2}{nc})=\frac{ab^2}{mnc^2}$$ Only true when $l=mnc$

Bernard
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April Jashmer Carba Anting
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