How many positive integer solutions satisfy this equation?

[songoku]

Hello pal @songoku, we can rewrite your second equation as
$$
x+y+1= \frac{xy}{2021}$$

Now, we want $x$ and $y$ to be integer, that implies $xy$ must be a multiple of 2021, right? So, we can write it as $xy =2021 k$.

Now, we can write out your third equation as:
$$
2021x + 2021y +2021= 2021k$$
Now, we have got a system:
$$
x+y =k-1$$

And $$xy =2021k$$
Taking $y=k$ and $x=2021$ would not satisfy, the first equation of our system, hence $x$ and $y$ must be the factors (or their multiples) of 2021. [Edited: or at most one of them can be a factor of $k$, but not k itself].

Hello @Hall

I got the answer. Thank you

To the OP: You have started correctly. Either your teacher taught you well, or he/she told you how to start. To continue, think about what you have been learning recently in pre-calc. Did you learn how to graph a hyperbola? Did you learn how to solve a degree-2 equation?

word

Yes, I know how to graph hyperbola from general equation:
$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$

But sorry I can't relate the question to hyperbola equation.

If by degree-2 equation you mean quadratic equation, then yes I can solve it using factorization, quadratic formula or completing square.

This question is given by the teacher only as a practice. He said this is our "holiday companion", we can do this to kill the time or just ignore it. So I am not really sure whether this question is related to what I have learned so far in the class.

Thanks

 

[OmCheeto]

Hello @Hall

I got the answer. Thank you

62?

 

[PeroK]

62?

Too many!

 

[OmCheeto]

Too many!

Wolfram Alpha and I both got 62. How many did you get?

[edit] Doh! It says "positive". Change that to 32.

 

[anuttarasammyak]

Now, we have got a system:
x+y=k−1

And xy=2021k

Applying familiar quadratic equation solution relation,
$$(X−x)(X−y)=X^2−(k−1)X+2021k=0$$
$$D=(k−1)^2−8084k=(k−4043)^2−(4043^2−1)>0$$
The problem is restated : how do we choose positive integer k so that
$$x,y=\frac{k−1±\sqrt{D}}{2}=2021+\frac{1}{2}(p±\sqrt{p^2−A})$$
where
$$p=k−4043$$
$$A=4043^2−1=4042∗4044=4∗2021∗2022=2^3∗3∗43∗47∗337=16,345,848$$ referring my previpus post, are positive integers and how many are such k ?

Say $p^2−A=q^2$ with positive integer q,
$$(p+q)(p−q)=A=2^3∗3∗43∗47∗337$$
There are 2^4 cases for which p+q or p-q is multiple of 3, 43, 47, 337.
There are 2^2 cases for such division of multiple 2^3, i.e. {1,8}{2,4}{4,2}{8,1}
So that both p+q and p-q are even, we should apply 2 cases of {2,4}{4,2} only.
So in total 2^5=32 x,y integer cases we have. Explicitly
$$x,y=2021+2^l 3^m 43^n 47^r 337^s, 2021+2^{1-l} 3^{1-m} 43^{1-n} 47^{1-r} 337^{1-s},$$
where
$$l,m,n,r,s=\{0,1\}$$

 

[Prof B]

62?

62 is the number of integer solutions. But the question asked for the number of positive integer solutions.

 

[Prof B]

Yes, I know how to graph hyperbola from general equation:
$$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$$

You get different kinds of equations depending on how the hyperbola is rotated. For example xy=1 is the equation for a hyperbola. It's easy to tell how many integer solutions that one has.

Your equation is also a hyperbola. If you write the equation in the appropriate form you'll find the asymptotes and also the number of integer solutions.

If by degree-2 equation you mean quadratic equation, then yes I can solve it using factorization, quadratic formula or completing square.

You also got a quadratic equation. It has two variables, so complete the rectangle instead of completing the square.

 

[songoku]

Thank you very much for the help and explanation Delta2, BvU, anuttarasammyak, WWGD, Hall, Prof B, OmCheeto, PeroK

 

[PeroK]

Thank you very much for the help and explanation Delta2, BvU, anuttarasammyak, WWGD, Hall, Prof B, OmCheeto, PeroK

We didn't see a lot of your work in this thread, it must be said!

 

[epenguin]

Hello @Hall

I got the answer. Thank you

We didn't see a lot of your work in this thread, it must be said!

Yup - don't tell us you got the answe, tell us the answer you got!

 

[songoku]

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{2021}$$
$$\frac{x+y+1}{xy}=\frac{1}{2021}$$
$$2021x+2021y+2021=xy$$
$$2021=xy-2021(x+y)$$
$$2021=(x-2021)(y-2021)-2021^2$$
$$(x-2021)(y-2021)=2021+2021^2$$
$$(x-2021)(y-2021)=2021 \times 2022$$

Let: $x-2021=a$ , then $y-2021=\frac{2021 \times 2022}{a}=\frac{43 \times 47 \times 2 \times 3 \times 337}{a}$

Since $a$ is integer then for $y$ to be integer, $a$ must be combination of factor of 43, 47, 2, 3, and 337.

Each number has 2 factors so the possible combinations for $a=2^5=32$

 

[Prof B]

There's some work left to do. If a is positive then x and y are greater than 2021 so x and y are positive. You get 32 solutions that way. What if a is negative? Why don't you get any more solutions that way?

 

[anuttarasammyak]

What if a is negative? Why don't you get any more solutions that way?

Obviously from post #35 $$x,y=2021-2^l 3^m 43^n 47^r 337^s, 2021-2^{1-l} 3^{1-m} 43^{1-n} 47^{1-r} 337^{1-s}$$
where
$$l,m,n,r,s=\{0,1\}$$ Another 32 solutions.

[EDIT] Now I find (x,y)=(0,-1),(-1,0) should be excluded thanks to @PeroK #44.

 

[PeroK]

There's some work left to do. If a is positive then x and y are greater than 2021 so x and y are positive. You get 32 solutions that way. What if a is negative? Why don't you get any more solutions that way?

First, assuming $a, b$ are positive. We have $x = a + 2021, \ y = b + 2021$ and $ab = 2021(2022)$.

We have the solution $a = 2021, b = 2022$ (and vice versa). Otherwise, $a > 2022, b < 2021$ or vice versa.

If we take the negative solutions $-a, -b$, then the first two solutions are invalid, as either $x$ or $y$ is zero. And the original equation involved reciprocals. Otherwise, we have $x > 0, y < 0$ or vice versa. So, there are no more positive solutions.

There are, however, $30 = 15 \times 2$ further solutions with one of $x, y$ negative and the other positive.