Wikipedia:Reference desk/Archives/Mathematics/2014 December 2

= December 2 =

Euler's theorem
Hello, I am reading Elementary number theory by Barton, section 7.3, Euler's theorem and have a mental block. It asks me to prove $$a^{37}=a\, (mod 1729)$$. I have verified this numerically so I know it's true. He gives me the hint that 1729=7.13.19. I get $$\phi(1729)=1296=36^2$$ so Euler's theorem gives me $$a^{36\times 36} = a (mod 1729)$$. So then $$(a^{36})^{37}=a^{37}$$ but there I get stuck. I have just found out that $$a^{36}$$ is not equal to 1 for any a except a=1. I don't seem to be able to extract anything about $$a^{37}$$. I am making very heavy weather of this, what am I missing? Robinh (talk) 04:10, 2 December 2014 (UTC)


 * You're using the hint wrong. What is relevant is that 37 mod φ(7) = 37 mod φ(13) = 37 mod φ(19) = 1, which means $$a^{37}\equiv a^{\phi(n)+1} \equiv a\pmod n$$, for n = 7, 13, and 19. (See the article on the Chinese remainder theorem if the significance of that is not immediately obvious to you.) --173.49.79.100 (talk) 05:56, 2 December 2014 (UTC)