# nth triangular number having same number factors as n

Find the number of $$n < 100$$ such that the nth triangular number has the same number of positive factors as $$n.$$

What I did: We know that the nth triangular number is $$\frac{n(n+1)}{2}$$ and I represented n with its general prime factorization $$n = p_1^{e_1} + p_1^{e_2} + ... + p_k^{e_k}$$ which means that it has $$(e_1 + 1)(e_2+1)...(e_k+1)~\text{factors}.$$ How would I proceed from here?

• You have to know 1) the formula for the $n$th triangular number, and 2) the formula for the number of "factors" of $n$, in terms of the prime factorization of $n$. But what exactly does "factors" mean? $12$ has six "divisors", three prime factors if you count with multiplicity, two prime factors if you don't. So which do you mean by "factors"? Oct 11, 2020 at 5:09
• I have clarified in my edits Oct 11, 2020 at 5:17

Let $$n = p_1^{k_1} \cdots p_m^{k_m}$$ be the prime factorization of $$n$$. It has $$(k_1+1)\cdots(k_m+1) \tag{*}$$ factors.

Note that $$n+1$$ and $$n$$ share no prime factors. So the prime factorization $$n+1 = q_1^{l_1} \cdots q_{m'}^{l_{m'}}$$ consists of primes $$q_i$$ that are distinct from the other primes $$p_i$$.

In particular, $$n(n+1) = p_1^{k_1} \cdots p_m^{k_m} q_1^{l_1} \cdots q_{m'}^{l_{m'}}$$ is the prime factorization of $$n(n+1)$$.

• If $$n$$ is even, then, $$p_1 = 2$$ and thus the number of factors of $$n(n-1)/2$$ is $$k_1 (k_2+1) \cdots (k_m+1) (l_1+1) \cdots (l_{m'}+1)$$. Equating this to ($$*$$) yields $$k_1+1 = k_1 (l_1+1) \cdots (l_{m'}+1)$$ which is only possible if $$k_1=1$$ and $$m'=1$$ and $$l_1=1$$ all hold, i.e. $$n+1$$ is prime and $$n$$ is not divisible by $$4$$.
• If $$n$$ is odd, then $$q_1=2$$, and thus the number of factors of $$n(n-1)/2$$ is $$(k_1+1)\cdots(k_m+1)l_1 (l_2+1) \cdots (l_{m'}+1)$$. Equating this to ($$*$$) forces $$l_1=1$$ and $$m'=1$$, i.e. $$n+1 = 2$$ or $$n=1$$.

In summary, the solutions (besides $$n=1$$) are any even $$n$$ that is not divisible by $$4$$ and such that $$n+1$$ is prime.

• Also $T(6) = 21$ has the same number of divisors as $6$ has. For even $n$ the condition becomes "$k_1 = 1$ and $n+1$ is prime", or "$n+1$ is a prime $\equiv 3 \pmod{4}$". Oct 11, 2020 at 12:45
• @DanielFischer Thanks for catching my error Oct 11, 2020 at 18:26
• @GerryMyerson Thanks for catching my error Oct 11, 2020 at 18:27