I'm working on showing non-trivial zero divisors exist in the Ring of Integers modulo $n^{2}$ , $n \in \mathbb{N}$ and as part of that I showed that
$a b = n^{2}$ where $a,b \in \mathbb{N}$
has a solution when $a = b = n$
and it got me wondering if this holds as the only solution in which this can occur?
So, Can a squared natural number be decomposed into the product of two distinct natural numbers?
Any tips/starting points would be greatly appreciated.
On
A solution for $ab=n^2$ where $a,b\in\mathbb N$, $a\neq b$ and $a,b\neq 1$ exists if and only if $n$ is not prime. This can easily be seen from the fundamental theorem of arithmetic.
On
My work on the problem has focused on generating Pythagorean triples as function of integer $x$ with solution that every integer $x$ ($x ≥3$) when squared (i.e. $x^2$) can be resolved into at least one pair of its distinct divisors ($j$, $k$; $j < k$) such that their product(s) represents $x^2$. The number of pairs is dependent on the type of integer $x$. For example, all odd primes ($j =1$, $k = x^2$) or twice odd primes ($j=2$, $k = \frac{x^2}{2}$) generate only a single such pair upon their squaring. There are worked out rules to predict pairs for every type of squared integer as per generalizations put forth in "Mir’s generalized theorem defining all integral Pythagorean triples as function of x" Shabir A. Mir : http://www.academia.edu/21914879/Mirs_generalized_theorem_defining_all_integral_Pythagorean_triples_as_function_of_x
Formulae and procedures have been listed therein. Lists generated with computational algorithm are also available therein.
Prof (Dr.) Shabir A. Mir 15-02-2019
If $n$ is of the form $$ n=p_{1}^{n_{1}}\cdot p_{2}^{n_{2}}\cdot \cdots \cdot p_{k}^{n_{k}},\qquad p_{i}\text{ prime, }1\leq i\leq k, $$ then a divisor $d$ of $n$ is of the form $$ d=p_{1}^{x_{1}}\cdot p_{2}^{x_{2}}\cdot \cdots \cdot p_{k}^{x_{k}},\qquad \text{ with }0\leq x_{i}\leq n_{i}\text{,} $$
Consequently, a divisor of $n^{2}$ is of the form $$ p_{1}^{y_{1}}\cdot p_{2}^{y_{2}}\cdot \cdots \cdot p_{k}^{y_{k}},\qquad 0\leq y_{i}\leq 2n_{i}, $$ which is one of the terms we obtain by expanding the product $$ \left( 1+p_{1}+p_{1}^{2}+\ldots +p_{1}^{2n_{1}}\right) \cdot \left( 1+p_{2}+p_{2}^{2}+\ldots +p_{2}^{2n_{2}}\right) \cdot \cdots \cdot \left( 1+p_{k}+p_{k}^{2}+\ldots +p_{k}^{2n_{k}}\right) . $$ This implies that the total number of divisors of $n^{2}$ is given by $$ \left( 2n_{1}+1\right) \cdot \left( 2n_{2}+1\right) \cdot \cdots \cdot \left( 2n_{k}+1\right) \geq 3. $$
The answer is affirmative when $n$ is such that $$ \left( 2n_{1}+1\right) \cdot \left( 2n_{2}+1\right) \cdot \cdots \cdot \left( 2n_{k}+1\right) >3. $$
For instance $$n^2=6^2=36=(2\cdot 3)^2=2^2\cdot 3^2,\qquad p_1=2,p_2=3,\quad n_1=n_2=1 $$ has $(2\cdot 1+1)\cdot (2\cdot 1+1)=9 $ divisors, while $$ n^2=3^2=9,\qquad p_1=3,\quad n_1=1 $$ has only $2\cdot 1+1=3 $ divisors.