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My Attempt is as follows:

If a is coprime to $35$, a must be coprime to $7$ and coprime to $5$.

Therefore by Fermat's little theorem, ${a}^{6}$gives a remainder of $1$ when divided by $7$, and ${a}^{4}$ gives a remainder of $1$ when divided by $5$.

${a}^{12}$ = ${7}\cdot {k} + {1}$ because of properties of congruent modulo

${a}^{12}$ = ${5}\cdot {m} + {1}$ because of properties of congruent modulo

since $7k=5m\;, 5|7k$ and $7|5m\;, k$ must be divisible by $5$ and $m$ must be divisible by $7$

${a}^{12}$ gives a remainder of $1$ when divisible by $35$.

Is my proof correct?

Jyrki Lahtonen
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Mathrocks123
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    Welcome to Mathematics Stack Exchange. Perhaps you should elaborate what you mean by “properties of congruent modulo” – J. W. Tanner May 15 '24 at 02:56
  • Do you know chinese remainder theorem? No matter, its essentially what you did. Your proof is correct but it's HUGE peeve of mine when people say things such as $a^{12} = 7\cdot k + 1$ without introducing $k$. I want to pin them in a corner and say "why? what is $k$. What if $k =19$? Are you really claiming that $a^{12} =134$" It's not really that hard to type "for same integer $k$", is it? But your proof is correct (but so would just citing the Chinese Remainder Theorem) – fleablood May 15 '24 at 05:00
  • Also here, if you prefer just $35$ :-) – Jyrki Lahtonen May 15 '24 at 05:44
  • For a solution-verification question to be on topic you must specify precisely which step in the proof you question, and why so. This site is not meant to be used as a proof checking machine. – Bill Dubuque May 15 '24 at 06:45

2 Answers2

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I like your solution. It's like, from scratch. That's almost always good.

To use a different approach, and a couple more advanced theorems, $\Bbb Z_{35}^×\cong \Bbb Z_7^××\Bbb Z_5^×\cong \Bbb Z_6×\Bbb Z_4\cong \Bbb Z_2×\Bbb Z_{12}.$

From this we easily see that $\Bbb Z_{35}^×$ has exponent $12.$

Succinctly, if $\lambda$ is Carmichael's function, then $\lambda (35)=12.$

suckling pig
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Your proof is correct. I'm going to offer an alternative approach.

If $a$ is coprime to $35$, then $a \equiv b \pmod {35}$, where $b \in U(35)$, the multiplicative group of units in $\Bbb Z / 35 \Bbb Z$, and $1 \leq b \leq 34$. Of course, $a^{12} \equiv b^{12} \pmod {35}$. Thus, we can work with $b$ without loss of generality.

It's easy to see that there are $24$ numbers less than $35$ that are coprime to $35$; i.e., $U(35)$ has order $24$. It's also easy to see that $U(35)$ isn't cyclic, because a cyclic group can have at most one element of order $2$, whereas $U(35)$ has at least $3$ such elements -- $34 (\equiv -1 \pmod {35}), 6, 29 (\equiv -6 \pmod {35})$.

By Sylow's Theorem, $U(35)$ must have an element $x$ of order $3$. It cannot also have an element $y$ of order $8$, because if such an element $y$ did exist, then $xy$ would have order $24$ and $U(35)$ would be cyclic after all. Thus, the only possible orders of elements of $U(35)$ are $1, 2, 3, 4, 6, 12$, so $\forall z \in U(35), z^{12}=1$.

Robert Shore
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