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

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?

• 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 $p = 2$. Dec 25, 2019 at 23:34
• This should be an answer. Dec 25, 2019 at 23:34

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 $$8$$.

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. $$\blacksquare$$

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.

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