Daniel's integer solutions

We factor $x^3y-xy^3$ $x^3y-xy^3=xy(x^2-y^2)=xy(x+y)(x-y)$ We have 3 terms being multiplied, but only two primes are being multiplied on the RHS. That means that one of the terms have to be 1. Now we do some case work:

1. if $xy=1$ : We see that it's not possible since x=1 and y=1 is the only solution that satisfies and $x-y=0$

2. if $x+y=1$ We see that this is also not possible since one of the terms will have to be 0

3. if $x-y=1$ We see that the only solution is $x=2$ and $y=1$ . (Other pairs won't work since they're being multiplied in $xy$ and again in $x+y$ meaning they will have more than two primes being multiplied.

So we see that the answers are $(2, 1, 2, 3)$ and $(2, 1, 3, 2)$ so $\boxed{2}$ pairs

$x^3y-xy^3=pq$ $xy(x^2-y^2)=pq$ $xy(x+y)(x-y)=pq$ Because $p$ and $q$ are primes and $LHS$ can be factored as a product of $4$ numbers, $2$ of them must equal $1$ . The only such possibility is when $x-y=1 \implies x=y+1$ Because $x$ and $y$ are positive integers, we know that $y=1, x=2$ . So, $pq=6$ Because $p$ and $q$ are primes, we have $\boxed{2}$ ordered quarduples, namely $(x, y, p, q)=(2, 1, 2, 3), (2, 1, 3, 2)$ .

To begin, we see that $x^3y - xy^3 = xy(x^2 - y^2) = xy(x - y)(x + y)$ .

This tells us that, if p and q are to be the only factors, exactly two of the multiplicands must equal 1. Since $x$ and $y$ are positive integers, the only options are $x$ , $y$ , and $x - y$ .

If $x = y = 1$ , then $x - y = 0$ , which does not satisfy the conditions. Similarly, if $x = x - y = 1$ , then $y = 0$ .That leaves us with $y = x - y = 1$ , meaning that $x = 2$ and $x + y = 3$ . These are both prime numbers, so the equality is satisfied.

That completely determines $x$ and $y$ . Since $p$ and $q$ can be assigned to either prime, we are left with $\boxed{2}$ answers.

Rewrite the given equation has $x\cdot y\cdot (x-y) \cdot (x+y)=pq$

Since RHS is the product of two primes, two of the factors (I don't know if it would be the right word) on LHS must be 1. It is clear that $x-y$ must be 1 and either of $x$ or $y$ must be 1 but since $x>y$ for the LHS to be positive, the only possible case is that $x-y$ and y must be 1.

When $y=1$ , $x-1=1 \Rightarrow x=2$ and $x+y=3$ . Hence, the only possible values for p or q are 2 and 3.

Therefore, the only possible quadruples are: $(2,1,2,3) \, \text{and} \, (2,1,3,2)$

Firstly, remark that p is a prime if and only if $p = 1 * p$

Next notice that $x^3y - xy^3$ is symmetric and hence the main idea here is to factorize the expression to take advantage of our observation at the start of the solution. We will first pull out a factor of $x$ and $y$ to get :

$x^3y - y^3x$ = $x * y * (x^2 - y^2)$

Notice that the expression can be further simplified using a difference of squares:

$x^3y - y^3x$ = $x * y * (x - y) * (x+y)$

Now we shall use the observation at the start of the solution. The observation implies that two terms in $x$ , $y$ , $x - y$ and $x + y$ are 1. Since $x$ and $y$ are postive integers , we see that $x + y > 1$ . Neither can both $x$ and $y$ be 1 or else the product of the terms will be 0 due to $x - y$ and 0 is not a prime. Due to the $x - y$ term, we get $x > y ≥ 1$ . Whence we can conclude that:

$y = 1$

$x - y = 1$

The second equation gives x = 2 . Notice that there is only one unique set of $(x,y)$ Subbing this back into the original equation we get:

$x^3y - y^3x$ = $1 * 1 *2 * 3$ .

so {p,q} = {2,3} , {3,2}

In conclusion, there are 2 solutions.

xy(x-y)(x+y) = pq

there are 2 factors in RHS and 4 in LHS. So two factors in LHS are 1. RHS is positive so LHS is also positive hence x>y so y=1 and x-y = 1

As x-1=1, x=2. LHS = 6=RHS. pq=6 p=3,q=2 and q=3,p=2

x=2,y=1,q=3,p=2 and x=2,y=1,p=3,q=2 are two set of solutions.

Factoring out, we get...

$\begin{aligned} &~ &x^3y-xy^3~~~&=~~~pq \\ &\Longrightarrow ~~~&xy(x^2-y^2)~~~ &=~~~ pq \\ &\Longrightarrow ~~~&xy(x+y)(x-y)~~~ &=~~~ pq \end{aligned}$

We have $4$ factors $x, y, (x+y), (x-y)$ on the left side and $2$ factors $p,q$ on the right side... Since $p$ and $q$ are prime numbers, they cannot be further factored into positive integers... So, we can conclude that exactly two of the factors on the left side are equal to $1$ and the other two are equal to $p$ and $q$ respectively...

Before going further, we should know that $x>y$ ... If $x=y$ , the LHS becomes $0$ and if $x<y$ , the LHS becomes negative, which is not possible as $p$ and $q$ are prime numbers...

Here, we can choose $2$ factors out of the $4$ factors on the left side in $\dbinom{4}{2}=6$ ways... That gives us $6$ cases... We'll keep in mind that $x$ and $y$ are positive integers while working with the cases...

Case 1: $x=y=1$ ... Then, $(x-y)=0$ , making the LHS become $0$ ... Hence, this case yields no solution...

Case 2: $x=(x+y)=1$ ... Then, $y=0$ , making the LHS become $0$ ... Hence, this case yields no solution...

Case 3: $x=(x-y)=1$ ... Then, $y=0$ , making the LHS become $0$ ... Hence this case yields no solution...

Case 4: $y=(x+y)=1$ ... Then, $x=0$ , making the LHS become $0$ ... Hence, this case yields no solution...

Case 5: $y=(x-y)=1$ ... Then, $x=2y$ ... Since, $y=1$ , $x=2\times 1 =2$ ... Now, this case can yield some solution as it follows our condition $x>y$ and doesn't make the LHS $~=0$ ... We'll come back to this later...

Case 6: $(x+y)=(x-y)=1$ ... Then, $y=0$ , making the LHS become $0$ ... Hence this case doesn't yield any solution too...

Whew! Now, the only case which can show some light is Case 5 , $y = (x-y) =1$ ... Hence, $x=2, y=1$ ... That clearly means that the other $2$ factors, $x$ and $(x+y)$ are equal to $p$ and $q$ in some order... We found out the values of $x$ and $y$ ... So, we get $x=2$ and $(x+y)=(2+1)=3$ as the only values of $p$ and $q$ ...

Hence, either $p=2, q=3$ or, $p=3, q=2$ ...

This gives us exactly $\Large{\fbox{2}}$ possible ordered quadruples... They are...

$\Large{(2,1,2,3)}$ and $\Large{(2,1,3,2)}$ !

$xy(x+y)(x-y)=pq$

Suppose $pq$ is odd, then all factors $x$ , $y$ , $x+y$ and $x-y$ have to be odd. But if $x$ and $y$ are odd, $x+y$ is even, so this leads to a contradiction.

Conclusion: $pq$ is even, and because $p$ and $q$ are prime, one of them is 2.(If $p=q=2$ , we get $(x)(y)(x^2-y^2)= 4$ which has no integer solutions.)

Assume for a moment that $p=2$ . We then get $(x)(y)(x+y)(x-y)= 2q$

Note that $x$ and $y$ cannot both be even, because $2q$ would be a multiple of 16. Nor can they both be odd, because $pq$ would have to be a multiple of 4, and p and q would have to be be 2.

The expression is positive, so $x-y>0$ and the largest factor is $x+y$ . Because $q$ is prime, it follows that $x+y=q, x-y=1, xy =2$ , and by adding and subtracting these we get $x=(q+1)/2$ $y=(q-1)/2$ $xy=2 \Rightarrow (q^2-1)/4=2 => q=3$

$x=2, y=1, p=2, q=3$

Earlier we chose to set $p=2$ , The alternative choice was to set $q=2$ , interchanging p and q.

So we found $\boxed{2}$ solutions

From the problem we know that LHS has two prime factor Factor. x³y - xy³ into xy(x-y)(x+y)

Since p,q is positive , x cannot be equal to y so x+y is at least 3 which has at least 1 prime factor Xy also has at least 1 prime factor because x > y ≥ 1 so x-y will have no prime factor (and must equal to 1)

As both (x+y) and xy has at least 1 prime factor And (x+y)(xy) has two prime factor We get that both x+y and xy is prime number

Substitute y = x-1 (x+y)(xy) = (2x-1)(x)(x-1) which tell we that x-1 must equal to 1 So the x is 2 and y is 1

The given equation can be written as $xy(x+y)(x-y)=pq$ . It shows that there are three factors. But the Right Hand Side (RHS) of the equation has only two prime numbers. Hence, $x-y=1$ , or, $x=y+1$ .

Thus, we get $(y+1)y(2y+1)=pq$ . Again there are three factors. This is only true when $y=1$ . So, $x=2$ . These values are fixed.

Therefore, $pq=6$ which shows $(p,q)=(2,3)$ or $(3,2)$ . So the ordered pairs are $(2,1,2,3)$ and $(2,1,3,2)$ .

Factorising we get x y (x+y) (x-y)

Since these are 4 factors and the RHS has two primes atleast two of these factors must be 1.

If p,q not equal to 2, then RHS is odd...then no solution exists as can be seen by considering

[ x even - y odd ],[ y even - x odd ,][ x even - y even ],[ x odd - y odd ] ,

In all these cases on of the LHS becomes even

Thus atleast one of p or q is 2 (the only even prime)

If p=q=2,then we find no solution as the smallest combination of x and y (2,1)

Gives LHS as 6 not 4

Thus one of p or q is 2 the other is an odd prime

Since RHS contains only one 2...either of x or y is 2 and the other is odd.....(INFERENCE)

LHS must also contain only one 2 as a factor

This is possible only for (2,1,2,3) and (2,1,3,2)

as higher values of x and y will give 2 odd factors and one factor as 2

We cannot have 2 odd factors for a single odd prime in The RHS