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I have to find the modular inverse of a sequence of numbers. When I do the inverse of $5\pmod {37}$, I get $-7$. $$37 = 7(5)+2$$ $$5 = 2(2)+1\text{, then}$$ $$2 = 1(37)-7(5).$$

so the inverse is $-7.$

But $-7\times 5 \pmod{37}$ is $2$. Shouldn't it be $1$?

I need to use this value for two later problems so it's messing it all up.

amWhy
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user48812
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1 Answers1

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Since $5$ is coprime to $37,$ we know by Bezout's identity that $5$ is invertible mod $37$. To compute the inverse one could employ the Extended Euclidean Algorithm (which can be implemented very conveniently by hand, see here). But that algorithm is a bit overkill for such small numbers. Instead, it's simpler to employ Gauss's Algorithm and some twiddling as follows

$$\rm mod\ 37\!:\,\ \frac{1}5 \equiv \frac{7}{35}\equiv \frac{-30}{-2}\equiv 15$$

Community
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Bill Dubuque
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  • I tried using the equations below (which should be the extended euclidean algorithm), but it doesn't seem to be working

    $$37=7(5)+2$$ $$5=2(2)+1$$ $$2=1(1)+1$$ then $$1=5−2(2)$$ $$1=5−2(37−7(5))$$ $$1=12(5)−2(37))$$

     – user48812 Nov 09 '12 at 18:22
  • Please try the convenient version of EEA that I linked to. It is generally easier to apply and almost always much less prone to errors. The correct back-substitution above is $$\rm 1 = 5 - 2(\color{#C00}2) = 5 - 2(\color{#C00}{37-7(5)}) = 5-2(-7)5 - 2\cdot 37 = (1!+2\cdot 7)\cdot 5-2\cdot 37$$ Therefore $\rm: 1 = 15\cdot 5 - 2\cdot 37: \Rightarrow\ 1\equiv 15\cdot 5\ , (mod\ 37)\ \ $ – Bill Dubuque Nov 09 '12 at 18:23