# Prove that the product of all the positive divisors of two numbers is equal implies the numbers themselves are equal.

Let $P(n)$ and $P(m)$ be the product of all the positive divisors of $n$ and $m$, two positive integers.

I know that $P(n) = n^{T(n)/2}$ and $P(m) = m^{T(m)/2}$, where $T(n)$ denotes the number of positive divisors of $n$. But then proving that $n = m$ from this is just as difficult as the original problem.. If I write $n$ and $m$ in their prime factored forms, I can conclude that for each $i$, we must have $T(n) k_i = T(m) s_i$, where $k_i$ and $s_i$ are the exponents of $p_i$ and $q_i$ in the prime factored forms of $n$ and $m$. But again, I have to prove that $T(n) = T(m)$. Could someone give me a proof for why the product being equal implies the number of divisors is equal?

Thanks.

• Have you taken combinatorics? – David Reed Nov 20 '17 at 5:55
• Other than highschool, no. – Saad Nov 20 '17 at 6:21
• Same question here. – Dietrich Burde May 11 '18 at 18:49

Because of the symmetry we can assume $n \ge m$. To look for a contradiction, assume $n \gt m$. We then have $n=m^{\frac {T(m)}{T(n)}}$ which says ${\frac {T(m)}{T(n)}} \gt 1.$ Let $\operatorname{ord}_p(n)$ denote the exponent of $p$ in the prime factorization of $n$. If you look at each prime that divides $m$ the same relationship exists:
$\operatorname{ord}_p(n)={\frac {T(m)}{T(n)}}\operatorname{ord}_p(m)$. But $T(m)$ is the product of $(\operatorname{ord}_p+1)$ over all the primes dividing $m$. Since the orders are greater for $n$, we have $T(n) \gt T(m)$, a contradiction. We conclude $n=m$.
• What does the notation $ord_p$ mean? – Saad Nov 20 '17 at 6:22