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$\begingroup$

We are working in: $ \mathbb{Z}_{103}$

Find x for which: $ \widehat{61}\cdot x = \widehat{1} $

When I learned how to solve equations like this my teacher told me to check all available values. That method is working well and fast when working with $ \mathbb{Z}_{n} $ with a n which is not bigger.

But in this case what can I do?

Thank You!!

PS: I know I should write the numbers with a "^" over the number but I don't know how to do that in MathJax. Please edit my question if necessarry. EDIT: Thank you Ross Millikan for \widehat{}!

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    You can write \hat {61} to get $\hat {61}$ I don't know how to get a wider hat. – 2017-01-22
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    Thank you very much @RossMillikan – 2017-01-22
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    I just found widehat , so you can do \widehat {61} to get $\widehat {61}$. Better yet. – 2017-01-22
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    Wow! Thanks a lot! I edited my question. – 2017-01-22

3 Answers 3

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Doing arithmetic modulo $\;103\;$ all through:

$$61=-42=-6\cdot7\;\;,\;\;\begin{cases}6\cdot17=102=-1\implies-6^{-1}=17\\{}\\7\cdot44=308=-1\implies 7^{-1}=-44=59\end{cases}$$

Thus finally

$$(-6)^{-1}\cdot7^{-1}=17\cdot59=\ldots$$

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More generally, you go through writing explicitly the Bezout identity.

If $(a,n)=1$ you know that there are integers $A$ and $B$ such that $$ aA+nB=1. $$ The integers $A$ and $B$ can be found by a (possibly repeated) application of the division algorithm.

In terms of congruences the identity displayed reads $$ aA\equiv1\bmod n $$ which is just what you want.

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${\rm mod}\ 103\!:\,\ \dfrac{1}{61}\overset{ \times\, 2}\equiv \dfrac{2}{19}\,\overset{\times\,5}\equiv\,\dfrac{10}{-8}\equiv\dfrac{5}{-4}\equiv\dfrac{108}{-4}\equiv-27\ $ by Gauss's algorithm.