Prove $ 2^{7}\cdot3^{7} \notin \{72, 108\}^{*} $

Question

I Want to prove $ 2^{7}\cdot3^{7} \notin \{72, 108\}^{*} $, where $A^* = \{72^n \cdot 108^m \mid n,m \in \mathbb(Z)_0 \land m+n > 0\}$. Can you please verify my proof?

Proof:

Let $ A = \{72, 108\} $ and assme $ 2^7\cdot3^7 \in A^*$. Therefore, there exist $ n, m \in \mathbb{N}_0, m + n \gt 0$, such that $ 2^7\cdot3^7 = 72^m\cdot108^n $.

The prime factorization of $ 72 $ is $ 2^3\cdot3^2 $ and the prime factorization of $ 108 $ is $ 2^2\cdot3^3 $.

Therefore, $ (2^2\cdot3^3)^n\cdot(3^2\cdot2^3)^m = 2^7\cdot3^7 $, which is the same as saying $ 3^{3n+2m}\cdot2^{3m + 2n} = 2^7\cdot3^7 $.

According to the Fundamental Theorem of Arithmetic, each composite number has a single and unique prime factorization. Hence:

$$ 3m + 2n = 7 $$ $$ 2m + 3n = 7 $$

We'll solve the system of equations.

$$ 3m + 2n = 7 \Rightarrow 2n = 7 - 3m \Rightarrow n = \frac{7 - 3m}{2} $$

Thus:

$$ 3n + 2m = 7 \Rightarrow 3[\frac{1}{2}(7 - 3m)] + 2m = 7 \Rightarrow 3(7 - 3m) + 4m = 14 \Rightarrow 21 - 9m + 4m = 14 \Rightarrow 21 - 5m = 14 \Rightarrow 5m = 7 \Rightarrow m = \frac{5}{7} \notin \mathbb{N}_0 $$

And that is a contradiction.

Answer 1

A slightly simpler version of your proof in the same vein. As noted, $72 = 2^33^2$ and $108 = 2^23^3$

Assume there exist integral $m, n$ such that $3m + 2n = 7$ and $2m + 3n = 7$. Then $5m + 5n = 14$ (by adding the two). Since $5$ does not divide $ 14$, a contradiction is reached. No such $m, n$ therefore exist.

Answer 2

Your argument is fine. It simplifies - to solve the equations simply add them, which immediately yields the modular contradiction: $\,\color{#c00}5$ divides $\color{#0a0}{14} \color{#c00}\Rightarrow\!\color{#0a0}\Leftarrow\:\!$. Quicker and simpler: compare the prime factorization lengths (#primes) on both sides of $\,72^{\rm N} 108^{\rm M} =2^7 3^7$ as below:

$$\begin{align}\text{ Suppose that}\ \ \overbrace{(2^{\large\color{#c00}3} 3^{\large\color{#c00}2})^{\rm N}(2^{\large\color{#c00}2} 3^{\large\color{#c00}3})^{\rm M}}^{\textstyle 72^{\rm N} \ 108^{\rm M}\!\!\!\!}\! &\,=\, 2^{\large\color{#0a0}7} 3^{\large\color{#0a0}7}\\[.2em] \text{$\color{#90f}{\rm count}$ #primes}\:\Rightarrow\: {\color{#c00}5\:\!{\rm\small N}\ \:\!+\:\!\ {\color{#c00}5\:\!{\rm\small M}}} &\,=\, \color{#0a0}{14}\ \color{#c00}\Rightarrow\!\color{#0a0}\Leftarrow\end{align}\qquad$$

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Generalization $ $ Define $\,\color{#90f}{\ell(n)}\,$ as the $\rm\color{#90f}{count}$ of all primes (length) in the unique prime factorization of $\:\!n.\,$ The above generalizes to: $\,\gcd(\ell(a),\ell(b))\mid \ell(a^n b^m)\ [= n\:\!\ell(a)+m\:\!\ell(b)].\,$ The question is the special case when $\, \ell(a) = \ell(b)\ [= 5]\, $ thus $\,\ell(a)\mid \ell(a^nb^m).\,$

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Remark $ $ Note this "global" method of counting the total number of primes ends up being simpler than the "local" methods of comparing the powers of each prime (e.g. using valuations $\,\nu_p).\,$ While generally the local methods can discriminate better, this extra work may not be needed to handle simple problems like this.

E.g. the global method quickly proves irrational square roots of odd length $n\,$ (e.g. prime $n)\,$ by $\,\sqrt n =a/b \Rightarrow nb^2 = a^2\,$ contra parity: $\,\ell(a^2)$ is even but $\,\ell(nb^2)\,$ is odd by $\,\ell(n)\,$ odd. But this method doesn't handle even length $\,n,\,$ so we need to resort to local methods to handle e.g. $\sqrt 6,\,$ e.g. by comparing the parity of the length (power) of $2$ on both sides (i.e. apply valuation $\nu_2)$.

This length "metric" $\,\ell(n)\,$ often proves handy, e.g. it can be used as a metric for descent in certain generalizations of the Euclidean algorithm to arbitrary PIDs, see the Dedekind Hasse PID test.

Notice that the proof uses only length uniqueness of prime factorizations. Domains satisfying this "half" of the common definition of unique prime factorization are known as half-factorial domains. A classical result of Carlitz shows that a number ring is half-factorial iff it has class number $\le 2.\,$ Recent work generalizes this, e.g. see here. One interesting simple result is: if a domain is not a UFD then that is witnessed by the existence of an equal-length nonunique factorization, see Coykendall and Smith, Unique factorization domains.

Answer 3

Once you figured out the particular form of the factorization of $72$ and $108$, there is a way to speedrun your exam.

In fact since $2^7\cdot 3^7$ involves very small exponents you can take advantage of this to enumerate the only $3$ possibilities that arise rather than going for a general arithmetic proof and save a lot of time to answer the other questions in the test.

$\begin{align}2^7\cdot 3^7&=54\times 72^2\\&=36\times 72\times 108\\&=24\times 108^2\end{align}$

Since all remainders $54,36,24$ are too small to constitute another factor of $\{72,108\}$ then we can safely conclude that such a factorization is not possible.