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I don't want the answer, but I need some help on how to figure out the answer. If you could point me in the direction of a useful math theorem or technique it would much much appreciated. Also, I am excluding 1 from the answer.

Clarinetist
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user3487324
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3 Answers3

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There is no $x\gt 1$ that works.

Hint: Let $a=3$.

André Nicolas
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  • 636 is a valid value for x. – Angad Jan 19 '15 at 03:25
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    @Angad: The answer above says it isn't. I don't want to go into detail since OP strongly specified she/he did not want a solution, only a hint. Nothing in the question says relatively prime. – André Nicolas Jan 19 '15 at 03:31
  • I am sorry, I must be missing your point. In my view though, the value of x does not matter. Just that some x may satisfy the condition is what matters. – Angad Jan 19 '15 at 03:32
  • There certainly are suitable $x$ if we want $a^x\equiv a \pmod{1926}$ for all $a$ in the interval such that $a$ and $1926$ are relatively prime. But OP did not say relatively prime. – André Nicolas Jan 19 '15 at 03:34
  • I see. I misread the question. I agree with your answer. – Angad Jan 19 '15 at 03:38
  • Maybe OP intended to specify relatively prime, but forgot to, in which case (without your recent change) your answer would be the useful one. – André Nicolas Jan 19 '15 at 03:42
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Since you have put conditions on the value of $a$ but $x$ is free. Then the smallest value of $x$ will be $1$.

Suppose if $x > 1$ and $a$ relatively prime to $1926$, then use Euler's theorem i.e. if gcd$(a, 1926) = 1$, then $a^{\phi(1926)} \equiv 1$ mod $1926$

user110219
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  • and then we find the phi of 1926, but that's a lot of numbers to list and we have to get rid of the numbers that are divisible. Like for example $\phi(6) \rightarrow {1,2,3,4,5,6}$ is 1 and 5 because $6 = 2 x 3$ and I can cross out 2,3,4, and 6. so that phi value is 2. But for this one, after we find the phi of 1926, then we need to find an a such that it can produce a remainder 1 in mod 1926 which is also a huge number to deal with. The only thing I can think of is that a needs to be 1. 1 in mod 1926 is 1. If we have 1927 then the result will also be 1 because we went through mod 1926 once. – usukidoll Jan 19 '15 at 10:06
  • I don't understand what do you mean. If we factorize $1926 = 29107$ and compute $\phi(1926)=636$. By theorem we know that $a^{636} \equiv 1$ mod $1926$ and multiplying both sides by a. – user110219 Jan 19 '15 at 14:24
  • I remember doing this problem before... but yes we can factorize 1926 and then find the $\phi(2),\phi(9), \phi(107) $. Multiplying those results we will have 636 and then through looking at the problem we can figure out the a which is only 1 because $1^{636} =1$ and that's the only way we can get the remainder 1 – usukidoll Jan 20 '15 at 04:13
  • $\phi(2) =1 , \phi(9) = 6,$ and since 107 is a prime we have $\phi(107) = 106$. By a Theorem: If p is a prime and a is a positive integer , then $\phi(p^a)=p^{a}-p^{a-1}$. Multiplying 1 x 6 x 106 gives us 636. So we have $a^{636} \equiv 1 mod 1926$ so we just need to figure out what value of a will give a remainder of 1 in mod 1926 – usukidoll Jan 20 '15 at 04:27
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Hint $\,\ n\mid a^e-a\,$ for all $\,a\,\Rightarrow\, n$ squarefree, $ $ else $\,p^2\mid e\mid p^e-p\,\Rightarrow\,p^2\mid p,\,$ by $\,e\ge 2$

Remark $\ $ Below is a more general result (Korselt's criterion for Carmichael numbers).

Theorem $\ $ For natural numbers $\rm\:a,e,n\:$ with $\rm\:e,n>1$

$\qquad\rm n\:|\:a^e-a\:$ for all $\rm\:a\:\iff n\:$ is squarefree, and prime $\rm\:p\:|\:n\:\Rightarrow\: p\!-\!1\:|\:e\!-\!1$

Proof $\ $ See this answer.

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Bill Dubuque
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