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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 $a^{37}=a\,(mod1729)$ . 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 $a^{36\times 36}=a(mod1729)$ . 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)[reply]

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)[reply]

Resolved