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Let $R = \mathbb Z_{40}$ and let $\odot$ be defined on $R$ as follows:

$$\begin{aligned} a \odot b = a + 25b-10 \end{aligned}$$

I need to check if this structure has an identity element, so:

$$\begin{aligned} a \odot \mathbb 1_{R} = a \end{aligned}$$ $$\begin{aligned} a +25e-10 = a \end{aligned}$$ $$\begin{aligned} 25e-10=0 \end{aligned}$$

how can I check for which $0 \leq e \leq 39$ values that equation is true? Is there an algorithm I can use?

Edit: first of all let me apologize because it turns out I am unfamiliar with a lot of concepts you've put in your answers, honestly I didn't quite understand everything, so I've taken out what I know I can handle (hopefully correctly) and I've dealt with it my way. So, almost all of you had suggested I had to solve the equation:

$$\begin{aligned} 25x \equiv_{40} 10 \end{aligned}$$

so as shown in the class I attended, I used the Euclidean method therefore I checked $gcd(25,40) \mid 10$ and after a semplification I got

$$\begin{aligned} 5x \equiv_{8} 2 \end{aligned}$$

as a few of you pointed out. All the solutions of the previous equation are the solutions to $5x \equiv_{8} 1 $, so I've determined $h,k \in \mathbb Z : 1 = 5h+8k$. After a few calculations I've found that $h = -3$ and $k = 2$ so the solutions to my original equation are all in

$$\begin{aligned} \{2+8k : k \in \mathbb Z\} = 2+8\mathbb Z \end{aligned}$$

What I don't get is how do I correctly get to the set of inverse elements, which is $\{2, 10, 18, 26, 34\}$, given the solution to the equation I've found? Is it enough (and correct) counting using modulo 8 starting from $k = 0$ while $2+8k \leq 39$?

haunted85
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6 Answers6

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\begin{align*} &25e -10 \equiv 0 \pmod{40} \\ \Leftrightarrow &40 | 25e-10 \\ \Leftrightarrow &8|5e-2\\ \Leftrightarrow &5e \equiv 2 \pmod 8 \end{align*} You have to find a multiplicatice inverse of $5$ modulo $8$. You can use the Euclidean algorithm or just guess to find $5\cdot5 = 25 \equiv 1 \pmod 8$. So $e \equiv 2\cdot 5 = 10 \equiv 2 \pmod 8$. The identity element $e$ is only defined up to multiples of 8, so when working modulo $40$, the elements $2,10,18,26,34$ are all identity elements.

marlu
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Hint $\rm\ \ 25\, e = 10+40\,k\iff 5\,e=2+8\,k\iff mod\ 8\!:\ e\,\equiv\,\dfrac{2}5\,\equiv\, \dfrac{10}5\,\equiv\, 2$

Thus $\rm\: e = 2 + \color{#0A0}8\,j.\:$ To get $\rm\:e\,$ mod $40 = \color{#0A0}8\cdot\color{#C00}{5},\:$ write $\rm\:j = \color{#C00}5\,q + r,\,$ for $\rm\,0\le r < \color{#C00}5.\,$ Hence $$\rm e\, =\, 2+8\,(r+5\,q)\, =\, 2 + 8\,r + 40\,q\, =\, \{2,10,18,26,34\} + 40\,q$$

See any decent textbook on elementary number theory for general methods to solve such linear diophantine equations, e.g. via the the extended Euclidean algorithm and Bezout identity for gcd. This topic is also discussed in numerous prior questions here (search using said buzzwords).

Bill Dubuque
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You want $25e - 10 \equiv 0 \mod 40$, or rather $25 e \equiv 10 \mod 40$. This is a linear congruence, so it's well understood. It turns out that since $\gcd (25, 40) = 5$, there will be $5$ solutions. One is $2$, as $50 - 40 = 10$. The others will be $10, 18, 26$, and $34$. (Check that this is always true: $250 - 240 = 10$, etc.)

For more on solving linear congruences, you might ask google or look through MSE for questions on linear congruences and/or modular arithmetic (e.g. General method for solving $ax\equiv b\pmod {n}$ without using extended Euclidean algorithm?).

Community
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davidlowryduda
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Work modulo 40:

$$25E-10=0\Longleftrightarrow 25e=10+40k\Longrightarrow 5e=2+8k=2\pmod 8$$ Since $\,5^{-1}=5\pmod 8\,$, (because $\,5\cdot 5=25=1+3\cdot 8\,$) , we get that $$e=5^{-1}\cdot 2=5\cdot 2=2\pmod 8$$ and thus $\,2\,$ is an answer. Can you take it from here?

DonAntonio
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Is there an algorithm I can use?

You are asking for the complete solution to

$$ 25 x - 10 \equiv 0 \pmod{40}. $$

Algebra still works in modular arithmetic, so we bring 10 over to the right hand side,

$$ 25 x \equiv 10 \pmod{40}, $$

and now we run into a tricky point: 25 is not invertible modulo 40, so we can't multiply through by an inverse of it as we normally would.

Because $\gcd(25,40) = 5$, it's easy to see that if 10 were a number not divisible by 5, there could be no solutions: reduce everything modulo 5. But because 10 is divisible by 5, we can cancel it to get the equivalent equation

$$ 5 x \equiv 2 \pmod 8 $$

5 is invertible modulo 8, so you can multiply through by its inverse. There are general algorithms for finding modular inverses, but in this case exhaustion is easy enough: $5 \cdot 5 = 1 \pmod 8$, and so we have equivalent equations

$$ 5 \cdot 5 x \equiv 5 \cdot 2 \pmod 8$$ $$ x \equiv 2 \pmod 8 $$

Each step was a reversible operation on equations, so this is the complete solution to your original equation: anything that reduces to 2 modulo 8 will satisfy $25 x \equiv 10 \pmod{40}$.

If we wanted to verify directly that all $x \equiv 2 \pmod 8$ are solutions:

$$\begin{align*} 25 x &\equiv25 \cdot 2 \pmod{8 \cdot 25} \\ &\equiv 50 \pmod{200} \\ &\equiv 50 \pmod{40} \\ &\equiv 10 \pmod{40} \end{align*}$$

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Here is a simple program line in GAP:

Filtered( [0..39], x -> RemInt(x*25,40)= 10 );

its result would be:

[ 2, 10, 18, 26, 34 ]

Mikasa
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  • This is simply an exhaustive search to solve the equation $25x \equiv 10 \pmod{40}.$ I'm sure you can do better in any computer algebra system e.g. here –  Jul 09 '12 at 16:33
  • @J.D.: Thanks for noting me that. Honestly, This program came to my mind suddenly and I just added it here. Other peoples solved the problem compeletly and I did nothing more. :) – Mikasa Jul 09 '12 at 16:37