Muhammad's Integers

Let $m=p_1^{a_1}p_2^{a_2}p_3^{a_3}p_4^{a_4}\ldots$ $n=p_1^{b_1}p_2^{b_2}p_3^{b_3}p_4^{b_4}\ldots$ where $p_1,p_2,p_3,p_4\ldots$ are prime numbers 2,3,5,7.....
$a_1,a_2,a_3,a_4\ldots,b_1,b_2,b_3,b_4\ldots$ are whole numbers.
Now \gcd(m^3,n^2)=p_1^\min(3a_1,2b_1)p_2^\min(3a_2,2b_2)p_3^\min(3a_3,2b_3)\ldots $\implies\min(3a_1,2b_1)=2$ Because $3a_1\neq2$ Therefore $2b_1=2\implies b_1=1$ Similarly $b_2=1$ , $\min(3a_3,2b_3)=0,min(3a_4,2b_4)=0\ldots$
Again lcm(m^2,n^3)=p_1^\max(2a_1,3b_1)p_2^\max(2a_2,3b_2)p_3^\max(2a_3,3b_3)\ldots $\implies\max(2a_1,3b_1)=2$ Because $3b_1\neq4$ Therefore $2a_1=4\implies a_1=2$ Similarly $a_2=2$ , $\max(2a_3,3b_3)=6,max(2a_4,3b_4)=0\ldots$
From above 2 results we get $a_4=a_5=a_6=\ldots=b_4=b_5=b_6=\ldots=0$ We also see that if: $a_3=3 \implies b_3=0$ and if $b_3=2 \implies a_3=0$
So we get that if:
$m=2^23^25^3$ than $n=2^13^1$ and if $m=2^23^2$ than $n=2^13^15^2$
Thus there are 2 ordered pairs (m,n) which satisfy the conditions.

Note that the prime factorization of $m, n$ must consists of only the primes $2, 3$ and $5$ . So, we can write $m = 2^{a} \cdot 3^{b} \cdot 5^{c}$ and $n = 2^{d} \cdot 3^{e} \cdot 5^{f}$ . Note that every set of $(a, b, c, d, e, f)$ determines a distinct solution, by the fundamental theorem of arithmetic.

Now, from $GCD(m^{3}, n^{2}) = 2^{2} \cdot 3^{2}$ , we get $min(3a, 2d) = 2 = min(3b, 2e)$ and one of $c$ or $f$ is $0$ . Then, it is obvious that $d = e = 1$ .

From $LCM(m^{2},n^{3}) = 2^{4} \cdot 3^{4} \cdot 5^{6}$ . Then, $a = 2, b = 2$ .

Now, we consider two cases.

1) $c = 0$

Then by $LCM(m^{2},n^{3}) = 2^{4} \cdot 3^{4} \cdot 5^{6}$ , one gets $f = 2$ .

2) $f = 0$

Then, $c = 3$ .

So, there are two solutions.

Clearly prime factors of m and n are 2 3 5 denote m=2^x1.3^x2.5^x3 and n=2^y1.3^y2.5^y3 then we can write gcd(m^3,n^2)=2^(min(3x1 and 2y1)).3^(min(3x2 and 2y2)).5^(min(3x3 and 2y3)) lcm(m^2,n^3)=2^(max( 2x1 and 3y1)).3^(max( 2x2 and 3y2)).5^(max(2x3 and 3y3)) => min(3x1 and 2y1)=2 but x1 and y1 are whole numbers so y1=1 lly max( 2x1 and 3y1)=4 so x1=2 IIY x2=2 y2=1 now min(3x3 and 2y3)=0 and max(2x3 and 3y3)=6 so(x3 y3)=(3 0) (0 2) correspond to solution (m n)=(4500 6) and(36 60) are two solutions

From $gcd(m^3,n^2) = 2^2 * 3^2$ follows that $n = 2 * 3 * X$ , since $n$ must contain the prime factors $2, 3$ but cannot contain each of them more than once as otherwise the second condition can not be fulfilled. It follows that $m = 2^2 * 3^2 * Y$ from the second condition. Moreover, we know that $5$ is not part of the gcd, so all prime factors of $5$ must be either contained completely in $n$ or completely in $m$ . This leaves two possibilities: $n = 2 * 3 ; m = 2^2 * 3^2 * 5^3$

$n = 5^2 * 2 * 3; m=2^2 * 3^2$ .

From the lcm result of $2^4 \cdot 3^4 \cdot 5^6$ , we can deduce that the prime factors of $m$ and $n$ can only be the subset of ${2, 3, 5}$ . Any other prime cannot be a factor of m or n , or else it would contribute to the result of the lcm.

Suppose $m = 2^{a} \cdot 3^{b} \cdot 5^{c}$ and $n = 2^{p} \cdot 3^{q} \cdot 5^{r}$ ...

There are a few facts can be derived from the equations:

The gcd result contains $2^2$ , which means $\min(3 a, 2 p) = 2$ ... this can only be achieved if $p = 1$ .

The gcd result contains $3^2$ , which means $\min(3 b, 2 q) = 2$ ... this can only be achieved if $q = 1$ .

The lcm result contains $2^4$ , which means $\max(2 a, 3 p) = 4$ ... this can only be achieved if $a = 2$ .

The lcm result contains $3^4$ , which means $\max(2 b, 3 q) = 4$ ... this can only be achieved if $b = 2$ .

Replacing the variables, we have so far:

$m = 2^2 \cdot 3^2 \cdot 5^{c} = 36 \cdot 5^{c}$

$n = 2^1 \cdot 3^1 \cdot 5^{r} = 6 \cdot 5^{r}$

Furthermore, $m$ and $n$ cannot be both a multiple of 5, or else the gcd result would include a multiple of 5.

Now if we want to calculate the exact pairs, we do the following:

If $m$ is a multiple of 5, then $c = 3$ and $r = 0$ , this is to make up the lcm result which includes $5^6$ , this leads to a pair of:

$m = 36 \cdot 5^{c} = 36 \cdot 5^3 = 36 \cdot 125 = 4500$

$n = 6$

If $n$ is a multiple of 5, then $r = 2$ and $c = 0$ , this is to make up the lcm result which includes $5^6$ , this leads to a pair of:

$m = 36$

$n = 6 \cdot 5^{r} = 6 \cdot 5^2 = 6 \cdot 25 = 150$

So the solution set is: $(4500, 6)$ and $(36, 150)$ ... 2 possible solutions.

We can also care less about the exact numbers. We are dealing with 6 variables, 4 of them $(a, b, p, q)$ are fixed to certain values. Out of the remaining variables, ... one of them will be zero, and the other non-zero, that is: $c = 0$ and $r \neq 0$ , or $c \neq 0$ , and $r = 0$ ... 2 solutions.

As $lcm(m^2,n^3)=2^4⋅3^4⋅5^6$ , let $m$ and $n$ be of the form $2^a⋅3^b⋅5^c$ and $2^d⋅3^e⋅5^f$ respectively. $(a,b,c,d,e,f \in \mathbb{N})$
$gcd(m^3,n^2)=2^2⋅3^2$ , so $min(3a,2d)=2 \Leftrightarrow$ either $3a$ or $2d$ equals $2$ . Because $m$ and $n$ are integers, clearly $3a>2d=2 \Rightarrow d=1$
Similarly, $3b>2e=2 \Rightarrow e=1$
Also, $min(3c,2f)=0 \Rightarrow c=0$ or $f=0$ $(*)$
On the other hand, for $lcm(m^2,n^3)=2^4⋅3^4⋅5^6$ , $max(2a,3d)=4$
Since $3d$ is already $3$ , it is obvious $2a=4$ or $a=2$ . Likewise, $b=2$
Again, $max(2c,3f)=6 \Leftrightarrow 2c=6>3f$ or $3f=6>2c$ . Together with the result $(*)$ above, we get two possible pairs of $(c,f)$ : $(3,0)$ and $(0,2)$
In conclusion, $2$ pairs of $(m,n)$ , which are $(2^2⋅3^2⋅5^3,2^1⋅3^1⋅5^0)$ and $(2^2⋅3^2⋅5^0,2^1⋅3^1⋅5^2)$ , satisfy the conditions.

Clearly, $m$ and $n$ each have some power of $2$ , some power of $3$ , and some power of $5$ in their prime factorizations. Let $m=2^a \cdot 3^b \cdot 5^c$ and $n=2^x \cdot 3^y \cdot 5^z$ , where $a,b,c,x,y,$ and $z$ are nonnegative integers. From $\gcd(2^{3a} \cdot 3^{3b} \cdot 5^{3c}, 2^{2x} \cdot 3^{2y} \cdot 5^{2z})=2^2 \cdot 3^2$ , we obtain that $\min(2^{3a}, 2^{2x})=2^2$ , so $x=1$ . We also get that $y=1$ from $\min(3^{3b}, 3^{2y})=3^2$ . And from $\min(5^{3c}, 5^{2z})=5^0$ , we get that at least one of $c$ and $z$ is $0$ .

From lcm $(2^{2a} \cdot 3^{2b} \cdot 5^{2c}, 2^{3x} \cdot 3^{3y} \cdot 5^{3z})= 2^4 \cdot 3^4 \cdot 5^6$ , we get that $\max(2^{2a}, 2^{3(1)})=2^4$ , so $a=2$ . Using similar reasoning, we get $b=2$ . From $\max(5^{2a}, 5^{3b})=5^6$ , we get that one of $2a$ and $3b$ is $0$ and the other is $6$ . There are $2$ choices so the answer is $2$ .

Since the greatest common divisor does not have a factor of $5$ , exactly one of $m^2$ or $n^3$ must have a factor of $5^6$ and the other has no factors of $5$ .

Suppose $n=5^2k$ . Then $\text{lcm}(m^2,k^3)=2^4 \cdot 3^4$ and $\gcd(m^3,k^2)=2^2 \cdot 3^2$ . Since the gcd contains a factor of 3, both $k$ and $m$ must contain a factor of $3$ . Furthermore, $k$ cannot contain a factor of $3^2$ , otherwise $\gcd(m^3,k^2)$ would contain a factor of $3^3$ . Using $\text{lcm}(m^2, k^3)=2^4 \cdot 3^4$ , we have $3^2$ is the largest factor of $3$ dividing $m$ . Similarly, $k$ contains a factor of $2$ , but not $2^2$ , and the largest factor of $2$ dividing $m$ is $2^2$ . Therefore, the only solution in this case is $m=2^2\cdot 3^2$ and $k=2\cdot 3$ .

Suppose $m=5^3k$ . Thus, $\text{lcm}(k^2, n^3)=2^4 \cdot 3^4$ and $\gcd(k^3, n^2)=2^2 \cdot 3^2$ . The argument above also applies to this case, since the conditions are the same with different variables. Then this case has exactly one solution $k=2^2\cdot 3^2$ and $n=2\cdot 3$ .

Therefore, there are $2$ possibilities.

From question above, we can conclude that there are just 3 prime factors: 2,3, and 5.

So, we can consider m= 2^a.3^a.5^b and n= 2^c.3^c or m= 2^a.3^a and n= 2^c.3^c.5^d.

With a= 2, b= 3, c= 2, and d= 2

Clearly, the prime factors of m and n are limited to $2, 3,$ and $5.$ Denote $m = {2}^{α_1} {3}^{α_2} {5}^{α_3}$ and $n = {2}^{β_1} {3}^{β_2} {5}^{β_3}$ . Then, $gcd( m^3 , n^2) = {2}^{min(3 α_1 , 2 β_1) } {3}^{min(3 α_2 , 2 β_2)} {5}^{min(3 α_3 , 2 β_3)}$

$lcm[ m^2 , n^3 ] = {2}^{ max(2 α_1 , 3 β_1)} {3}^{max(2 α_2 , 3 β_2)} {5}^{max(2 α_3 , 3 β_3)} .$

So, $min(3 α_1 , 2 β_1) = 2.$ But $α_1 and β_1$ are whole numbers, so it follows that $β_1 = 1.$ Similarly, $max(2 α_1 , 3 β_1) = 4, so α_1 = 2.$

We may similarly conclude that $β_2 = 1 and α_2 = 2.$ Now, $min(3 α_3 , 2 β_3) = 0 and max(2 α_3 , 3 β_3) = 6, so (α_3, β_3) = (3 , 0) or (0 , 2).$ These infer that $(m, n) = (4500, 6),(36, 60) .$

Hence $\boxed{2}$ solutions.

First, we define $gcd(x,y)$ and $lcm(x,y)$ in terms of the prime factorizations of $x$ and $y$ .

I created a system that follows the logic of least common multiples and greatest common denominators. I don't believe it follows standard set operations, but I found it quite useful. In this method, there may be repeated values in a set. With union operations, the greater number of times the repeat term appears is the number of times it will appear in the result (ex. $\left\{ 2,2,2,4,5,5,6 \right\} \cup \left\{ 2,2,4,4,7 \right\} =\left\{ 2,2,2,4,4,5,5,6,7 \right\}$ ). With intersection operations, the lesser number of times the repeat term appears is the number of times it will appear in the result (ex. $\left\{ 2,2,2,4,5,5,6 \right\} \cap \left\{ 2,2,4,4,7 \right\} =\left\{ 2,2,4 \right\}$ ). Also, I use $\sim { P }_{ a }$ to represent the values not contained in the set ${P}_{a}$ , where repeated terms repeated more often in the actual set than the "not" set are all moved to the result (ex. $\left\{ 2,2,4,4,7 \right\} \cap \sim \left\{ 2,2,2,4,5,5,6 \right\} =\left\{ 4,4,7 \right\}$ ).

Let ${ P }_{ x }$ be the set that contains the prime numbers that evenly divide an integer $x$ . I will call this the prime factor set of $x$ . If a prime factorization requires repeated digits, repeat values in the set. The greatest common factor of two numbers $x$ and $y$ will be equivalent to the number whose prime factor set is ${ P }_{ x }\cap { P }_{ y }$ . In other words, once we create this new set, we will find the product of its values.

The least common multiple, then, of the two numbers $x$ and $y$ is the product of the terms of $\left( \sim { P }_{ x }\cap { P }_{ y } \right) \cup { P }_{ x }$ .

Now we look at the first given, $gcd\left( { m }^{ 3 },{ n }^{ 2 } \right) ={ 2 }^{ 2 }\cdot { 3 }^{ 2 }$ . This means that in our previously defined method, ${P}_{m^3}\cap {P}_{n^2}=\left\{ 2,2,3,3 \right\}$ . Since the intersection operator requires the resulting values to be present in both $m^3$ and $n^2$ , we know that both ${P}_{m}$ and ${P}_{n}$ must contain at least one $2$ and one $3$ . However, this makes ${P}_{m^3}$ already have $3$ of each term, so we know that we must use ${P}_{n^2}$ to cap the number of repeats at two. This restricts the value of ${P}_{n}$ to exactly one $2$ and one $3$ .

Next, we examine the second bit of information, that $lcm\left( { m }^{ 2 },{ n }^{ 3 } \right) ={ 2 }^{ 4 }\cdot { 3 }^{ 4 }\cdot { 5 }^{ 6 }$ . We have already determined that ${P}_{n}$ contains exactly one $2$ and one $3$ (corresponding to three of each in ${P}_{n^3}$ ), so by our definitions of sets in union, the $2^4$ and $3^4$ must come from ${P}_{m^2}$ . This means ${P}_{m}$ must contain exactly two $2$ s and two $3$ s.

We still must account for the $5^6$ that remains in the second bit of information. It may be contained in ${P}_{n^3}$ or ${P}_{m^2}$ , but not both. If a $5$ appeared in both in any quantity, then the greatest common divisor would not equal $2^2\cdot 3^2$ . Note that if the exponent on $m$ or $n$ did not evenly divide the exponent of $5$ , we would have to eliminate one or both possibilities. Since both the exponent of $m$ and the exponent of $n$ evenly divide the exponent of $5$ , though, we can say there are two possibilities.

We know there are not more than two possibilities. If $m$ or $n$ had a larger exponent of any prime number (or a new prime number altogether) in their prime factorizations, they would affect the least common multiple or greatest common factor outside what we have been told to account for.

Because of $gcd(m^3,n^2) = 2^2 . 3^2$ , it follows that for some integers $q_{m}, q_{n}$ :

$m^3 = q_{m}. 2^2 . 3^2 \wedge n^2 = q_{n} . 2^2 . 3^2 .$

Because both m and n are integers, we deduce that the prime factorisation of $q_{n}$ cannot contain 2 or 3.

Let us suppose it would contain 2 (same argument holds for 3). Because m is an integer, the prime factorisation of $q_{m}$ has to contain 2, since the gcd is taken of $m^3$ and $n^2$ . If $q_{m}$ didn't contain 2, m wouldn't be an integer.

This however implies the gcd would contain another factor 2, so it is not possible.

Now looking at the lcm, and because only m can contain more factors 2 and 3 to make the lcm work, we conclude that :

$m = q'_{m}. 2^2 . 3^2 \wedge n = q'_{n} . 2. 3 ,$

for some integers $q'_{m}, q'_{n}$ , which now both don't contain any factors 2 or 3 anymore.

We only need to figure out what the fives can still do. They are in $q'_{m}$ and $q'_{n}$ .

There are only two possibilities with the given powers of m and n in the lcm : $(5^0, 5^2)$ and $(5^3, 5^0)$ for $q'_{m}, q'_{n}$ respectively.