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Recently I thought about the number $8$.

Is $8$ the only number with four divisors, the second largest of which is even?

$8$ certainly is such a number, since its divisors are $1, 2, 4$, and $8$ itself. If it is the only one, how can I prove it?

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    $\begingroup$ The only possibilities are $pq$ and $p^3$ for any primes $p$ or $q$, the second largest for the first case can't be divisible by $2$, as it is the larger of the primes, and $p^3$ only works for $p = 2$. $\endgroup$
    – ETS1331
    Dec 25, 2019 at 23:34
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    $\begingroup$ This should be an answer. $\endgroup$ Dec 25, 2019 at 23:34

3 Answers 3

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It is true.

For a number to have exactly four divisors, we require that it is either $p^3$ with $p$ prime (case 1) or else $qp$, with $p, q$ different primes (case 2).

However, as its second largest divisor is even, the number is even. Thus, in the first case the only possible prime is $2$, giving us the number $8$, and in case $2$ we have $2p$, with $p>2$. But then its second largest divisor will be $p$.

Hence the only number with your property is $8$.

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Take an $n$ with the aforementioned properties. Since $n$ has an even divisor, it must be divisible by two. Then, for its second largest divisor $\frac n2$ to be even, $n$ must be a multiple of $4$.

Now, suppose that there existed some prime $p\neq2$ that divided $n$. This would mean that $$1,2,4,p,2p,4p$$ are six different divisors of $n$, which is impossible. Therefore, $n$ must be a power of $2$. However, $2^k$ has $k+1$ divisors $$1,2,4,\ldots,2^k.$$ The only possibility is therefore that $k=3$ and $n=8$. That is, $n=8$ is the only number with your properties. $\blacksquare$

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Every positive integer has a unique factorisation into the form $p_1^{\alpha_1}p_2^{\alpha_2}\dotsm p_k^{\alpha_k}$ for primes $p_i$ and nonnegative integers $\alpha_i$. The number of divisors of such a number is $(\alpha_1+1)(\alpha_2+1)\dotsm(\alpha_k+1)$, so in order for a number to have exactly $4$ distinct divisors, then it must be either of the form $p^3$ for some prime $p$, or the form $pq$ for distinct primes $p$ and $q$. Now, since your second largest divisor is even, then the number itself must be even, so that $2$ must be one of the primes dividing it. In the first case this gives rise to $2^3=8$ as you observed. In the second case, the second largest divisor of an integer of the form $2q$ must be $q$, but since $q\neq2$ is prime then it must be odd. So you are right, $8$ is the only such number.

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  • $\begingroup$ When you said “$p$ must be one of the primes dividing it” did you mean $2$ must be? $\endgroup$ Dec 25, 2019 at 23:46
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    $\begingroup$ @J.W.Tanner Of course, thanks for catching the mistake. $\endgroup$
    – YiFan Tey
    Dec 25, 2019 at 23:46

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