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The problem comes from a practice final for a final exam I have later today.

It says "Show that if $\gcd(e, 24) = 1$ then $e^2 \equiv 1 \bmod 24$".

I found that Euler's totient function $\phi(24) = 8$ So I now know $e^8 \equiv 1 \bmod 24$, but I'm not sure where to go from there.

I found that if $\sqrt[4]e$ is an integer, then it's obvious that $\sqrt[4]e \mid e$, so $\gcd(\sqrt[4]e, 24) = 1$ which I can use to prove $e^2 \equiv 1 \bmod 24$, but that only proves it in the case where $\sqrt[4]e$ is an integer (and I don't think I'm really going in the right direction here).

Paul
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4 Answers4

16

Euler's $\varphi$-function is often not the right tool for this kind of problem. I would work separately modulo $3$ and modulo $8$.

If $\gcd(e,24)=1$, then $\gcd(e,3)=1$. Therefore, by Fermat's Theorem (but that's overkill!) we have $e^2\equiv 1\pmod 3$. It is overkill because if $e$ is not divisible by $3$, then $e\equiv \pm 1\pmod{3}$, and therefore $e^2\equiv 1\pmod 3$.

If $\gcd(e,24)=1$, then $e$ is odd. It is a standard fact that if $e$ is odd, then $e^2\equiv 1\pmod 8$. For a low level proof, all we need to do is to check the result for $e=1$, $3$, $5$, and $7$, or more simply for $e=\pm 1$ and $e=\pm 3$. Or else we can note that if $e$ is odd, then $e=2k+1$ for some $k$. Thus $e^2=4k^2+4k+1=4k(k+1)+1$. Since $k$ and $k+1$ are consecutive integers, one of them is even, and therefore $4k(k+1)$ is divisible by $8$.

From the facts that $e^2\equiv 1\pmod 3$ and $e^2\equiv 1\pmod 8$, we conclude that $e^2\equiv 1\pmod{24}$.

André Nicolas
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    To see that $e^2\equiv 1\pmod 8$ for $e$ odd, you can do the following simple argument: $e^2-1=(e-1)(e+1)$ is the product of two consecutive even numbers. Thus one must be a multiple of $4$ and the other is even.... – N. S. Dec 17 '11 at 19:23
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The answer by Andre Nicolas is the way to go. However always bear in mind that you could have just checked this by hand, because the modulus of the problem ($24$) is rather small!

If $\gcd(e,24)=1$, and we are going to compute a value modulo $24$, then it suffices to check that the statement is true for those all congruence classes $e\bmod 24$ such that $\gcd(e,24)=1$, i.e., we need to check the statement for $$e\equiv 1,5,7,11,13,17,19,23 \bmod 24.$$ Now the problem has been reduced to checking that the square of each one of these eight numbers is congruent to $1\bmod 24$. Indeed: $$1^2\equiv 1,\ 5^2\equiv 25\equiv 24+1\equiv 1,\ 7^2\equiv 49\equiv 48+1\equiv 1,\ 11^2\equiv 121 \equiv 24\cdot 5+1\equiv 1 \bmod 24,$$ and $$23^2\equiv (-1)^2\equiv 1^2\equiv 1,\ 19^2\equiv (-5)^2\equiv 5^2\equiv 1,\ 17^2\equiv (-7)^2\equiv 7^2\equiv 1,\ 13^2\equiv (-11)^2 \equiv 1 \bmod 24.$$ Thus, $e^2\equiv 1 \bmod 24$ whenever $\gcd(e,24)=1$.

Álvaro Lozano-Robledo
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Say $gcd(e, 24) = 1$. Since $24 = 2^3 \times 3$, we know $e$ is not even and not a multiple of 3. So $e$ is of form $6k \pm 1$, for some integer $k$.

Then $$(6k \pm 1)^2 = 36k^2 \pm 12k + 1$$ and so it suffices to show that $36k^2 + 12k$ is a multiple of $24$. We can factor it as $12k(3k+1)$; for any choice of $k$, one of $k$ and $3k+1$ is even, so $k(3k+1)$ is even, and $12k(3k+1)$ is a multiple of 24.

Michael Lugo
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A more sophisticated answer is that $U_{24} \cong U_8 \times U_3 \cong C_2 \times C_2 \times C_2$, and so has exponent $2$.

In general, the exponent of $U_m$ is $\lambda(m)$, where $\lambda$ is Carmichael's function.

lhf
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