# Find all N in $\phi(N)=98$ [closed]

Solve the equation $\phi(N)=98$

I have no idea how to do it. How to find all N?

## closed as off-topic by Watson, C. Falcon, choco_addicted, Dietrich Burde, user91500Jun 21 '16 at 12:58

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• Do you know how to use the prime factorisation of $N$ to calculate $\phi(N)$? – Arthur Jun 21 '16 at 12:11
• If $n=p^aq^b\dots$ for distinct primes $p,q,\dots$ then $\phi(n)=n(1-\frac{1}{p})(1-\frac{1}{q})\dots$. You know that $98=2\cdot7\cdot7$. So just look at the possibilities. $p-1$ must divide $98$ and so must $p$ if the power is greater than 1. – almagest Jun 21 '16 at 12:11
• Here, I found them: $\{\;\}$. – Ivan Neretin Jun 21 '16 at 12:14
• In any case, solving $\phi(n)=a$ is a finite problem because $\phi(n) \geq \frac{\sqrt{n}}{\sqrt{2}}$ and so you only need to test for $n\le 2a^2$. See math.stackexchange.com/questions/301837/…. – lhf Jun 21 '16 at 12:42

A prime factor $p$ of $N$ contributes a factor $p-1$ to $\phi(N)$ and also a factor $p^{e-1}$ if $p^e$ divides $N$.

Since $98 = 2 \cdot 7 \cdot 7$, consider where $7$ comes from.

It cannot come from $p=7$, because $6$ does not divide $98$.

It cannot come from $p-1$, because $8$ is not prime.

It must come from a factor of $p-1$.

The two factors of $7$ in $98$ cannot come from the same $p-1$ because neither $50$ nor $99$ are primes.

But if the two factors of $7$ in $98$ came from different $p-1$, we would have a factor of $4$ in $98$.

Bottom line, $\phi(N)=98$ has no solutions.

• But 7 might also be a non-trivial factor of $p-1$. – almagest Jun 21 '16 at 12:20
• @almagest, thanks. I think I've fixed it now. – lhf Jun 21 '16 at 12:36

Clearly $N\ge 2$. If $N=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_n^{\alpha_n}$ is the unique prime factorization of $N$ (with $p_1<p_2<\cdots <p_n$), then $$\phi(N)=p_1^{\alpha_1-1}p_2^{\alpha_2-1}\cdots p_n^{\alpha_n-1}\times$$

$$\times (p_1-1)(p_2-1)\cdots (p_n-1)=2\cdot 7^2$$

If $N$ is even, then $p_1=2$ and $$2^{\alpha_1-1}p_2^{\alpha_2-1}\cdots p_n^{\alpha_n-1}\times$$

$$\times (p_2-1)\cdots (p_n-1)=2\cdot 7^2,$$

and $p_2-1$, $\cdots$, $p_n-1$ are all even, so there are two cases:

$a)$ Here $\alpha_1=2$ and $n=1$, but then $\phi(N)=2\neq 2\cdot 7^2$.

$b)$ Here $\alpha_1=1$ and $n=2$. Then $$\phi(N)=p_2^{\alpha_2-1}(p_2-1)=2\cdot 7^2$$

$p_2$ is an odd prime, so we have two cases:

$a)$ Here $p_2=7$, but then $3\mid p_2-1$, so $\phi(N)\neq 2\cdot 7^2$.

$b)$ Here $\alpha_2=1$. Then $p_2=99$ is not prime.

If $N$ is odd, then, since $p_1-1$, $p_2-1$, $\ldots$, $p_n-1$ are all even, we get $n=1$, so $$\phi(N)=p_1^{\alpha_1-1}(p_1-1)=2\cdot 7^2$$

We get the same situation as before (with two cases). No solutions.

Note that $\phi(n)$ is even for all $n\gt2$. Consequently, if $2\lt m,n$ with $\gcd(m,n)=1$, then $4\mid\phi(mn)$, since $\phi(mn)=\phi(m)\phi(n)$. Now $4\not\mid98$, so $\phi(N)=98$ is possible only if $N=p^r$ or $2p^r$ for some odd prime $p$. But $\phi(2p^r)=\phi(p^r)=p^{r-1}(p-1)=98=2\cdot7^2$ clearly has no solution, mainly because $p=99$ is not a prime. (More precisely, if $r=1$ then $p=99$, while if $r\gt1$ then $p=7$, in which case $p-1=6$ implies $3\mid\phi(N)$, each of which is a contradiction.)