How many positive integer solutions satisfy this equation?

[songoku]

Homework Statement

How many positive integer solutions (x, y) satisfy $\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{2021}$

Relevant Equations

Not sure

$$\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}=\frac{1}{2021}$$
$$\frac{y+x+1}{xy}=\frac{1}{2021}$$
$$xy = 2021y + 2021 x + 2021$$

Then I am stuck. How to continue?

Thanks

 

[PeroK]

Did the question of whether 2021 is prime occur to you?

 

[songoku]

Did the question of whether 2021 is prime occur to you?

No

2021 is not prime (2021 = 43 x 47) but sorry I don't know how to use the hint to continue. To be honest, idea of using prime number to solve this question never crosses my mind.

Thanks

 

[PeroK]

To be honest, idea of using prime number to solve this question never crosses my mind.

That's a fundamental problem. It's like a golfer who cannot figure out how to get out of a bunker and it never crosses their mind to use a sand wedge! Or a chef who can't light a stove and never thinks of using a match!

We cannot help you to think for yourself. Primes are so fundamental to number theory that it's impossible for me to understand why you didn't think of the prime factorisation of 2021.

 

[songoku]

That's a fundamental problem. It's like a golfer who cannot figure out how to get out of a bunker and it never crosses their mind to use a sand wedge! Or a chef who can't light a stove and never thinks of using a match!

We cannot help you to think for yourself. Primes are so fundamental to number theory that it's impossible for me to understand why you didn't think of the prime factorisation of 2021.

I never learn about number theory. This question is a challenge question from the teacher, just for practice and optional for the students to do it or not.

What do I need to learn and understand to be able to do this question?

Thanks

 

[PeroK]

What do I need to learn and understand to be able to do this question?

I don't know. I don't know what you've studied and what you haven't and why you are doing these problems.

The last problem was the same. You had a circle and a point. My first thought was to use geometry and symmetry before diving into the algebra. You dived straight into the algebra. Why do I think of geometry in such a problem and you don't? I can't explain that.

 

[songoku]

Thank you very much PeroK

 

[Delta2]

Hmm I can't solve this, though I probably have 25-30 years more experience than @songoku

@PeroK if I understand your hint, x and y must relate somehow to 43 and 47 but i can't think exactly how, you got to give us a secondary hint

 

[PeroK]

Hmm I can't solve this, though I probably have 25-30 years more experience than @songoku

@PeroK if I understand your hint, x and y must relate somehow to 43 and 47 but i can't think exactly how, you got to give us a secondary hint

I haven't looked at it other than to check whether 2021 was prime. It's clear, therefore, that $x$ and/or $y$ must have factors of $43$ and/or $47$.

It might still be difficult, but that must be the first thing to look at.

 

[PeroK]

Here's one solution: $x = 16512, \ y = 2303$.

 

[Delta2]

Hmmm, did you use a python program to get those?

 

[PeroK]

Hmmm, did you use a python program to get those?

No. Pen and some Pret-a-Manger napkins!

 

[PeroK]

I think I've found them all. Not easy, but ultimately all about prime factorisation!

 

[anuttarasammyak]

$$(\frac{1}{x}+1)(\frac{1}{y}+1)=\frac{1}{2021}+1$$
$$\frac{(x+1)(y+1)}{xy}=\frac{2022}{2021}=\frac{2*3*337}{43*47}$$
Is this helpful ?

 

[BvU]

Does it help to "discover" that $${1\over 2z} + {1\over 2z+1} + {1\over 2z(2z+1)} = {1\over z}\qquad ?$$(for which I needed to replace the 2021 by 4 and do some trial-and-error, to find 8,9 does the trick. But then so does 6,14 and I still have to find out how and why...)

 

[PeroK]

Is this helpful ?

Very!

 

[PeroK]

To give some hints for what I did (which is not necessarily the quickest way);

There were two cases: a) $x = 2021k$ for some integer $k$; and, b) $x = 43k, y = 47m$ for some $k, m$.

a) leads quickly to a condition on $k - 1$.

For b) I showed first that $k > 47$ and $m > 43$.

 

[Hall]

As @songoku has mentioned that he has not been confronted (if that’s the right word here) with these types of problems before, I think we should at least guide him in a direction which would let him handle these types of problems instead of scathing him for why he didn’t do what he ought to.

 

[Hall]

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].

 

[PeroK]

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.

Well, $x = 4042, y = 4043$ is a solution. So, your conclusion that $y$ must be a multiple of the factors of $2021$ is false.

 

[Hall]

Well, $x = 4042, y = 4043$ is a solution. So, your conclusion that $y$ must be a multiple of the factors of $2021$ is false.

A nice counter-one. I have edited my post.

 

[PeroK]

A nice counter-one. I have edited my post.

More formally, $43$ divides $xy$, hence $43$ divides $x$ or $43$ divides $y$. Similarly, $47$ divides $x$ or $47$ divides $y$.

This gives us four possibilities, which reduces to two by the symmetry of $x$ and $y$:

a) $2021$ divides $x$ (or, $2021$ divides $y$)

b) $43$ divides $x$ and $47$ divides $y$ (or, vice versa).

Which is expanded on in post #17.

 

[Hall]

@PeroK It's so charming that there is no solution for first 2021 values of $k$!

 

[WWGD]

Hmm I can't solve this, though I probably have 25-30 years more experience than @songoku

@PeroK if I understand your hint, x and y must relate somehow to 43 and 47 but i can't think exactly how, you got to give us a secondary hint

Edit:This is not a solution, because I can't think of one, but a suggestion for experiments:
Notice the function is symmetric in x,y, i.e., f(x,y)=f(y,x).
0)Test to see what happens when
0.1)Both x,y are small
0.2)Both are large
0.3)x is significantly larger
0.4) x=y ; x=y=2021, or x=y =k2021; k an integer.

1)Go to Wolfram and plot the function. Find it's max and min. Maybe there's a way of restricting to Integers.

2)Look for other interpretations. This is close to being the ratio of the perimeter of a rectangle to its area.

Hopefully this brainstorming will lead to an insight. Edit: Tl; dr: There is no royal road to Geometry, nor to any advanced Mathematics. Need to explore, struggle until something falls through.

 

[anuttarasammyak]

From my post #14 I would rewrite the problem as

Find x,y satisfying
$$Pa=Rc-1\equiv x$$
$$Qb=Sd-1\equiv y$$
$$ab=cd$$
where a,b,c,d are positive integers,
$$\{P,Q\}=\{1,43*47\},\{43,47\},\{47,43\},\{43*47,1\}$$ of four cases, and
$$\{R,S\}=\{1,2*3*337\},\{2,3*337\},\{3,2*337\},..., \{2*3*337,1\}$$ of eight cases.

I am not sure such a rewriting would be useful.

 

[Hall]

Edit:This is not a solution, because I can't think of one, but a suggestion for experiments:
Notice the function is symmetric in x,y, i.e., f(x,y)=f(y,x).
0)Test to see what happens when
0.1)Both x,y are small
0.2)Both are large
0.3)x is significantly larger
0.4) x=y ; x=y=2021, or x=y =k2021; k an integer.

1)Go to Wolfram and plot the function. Find it's max and min. Maybe there's a way of restricting to Integers.

2)Look for other interpretations. This is close to being the ratio of the perimeter of a rectangle to its area.

Hopefully this brainstorming will lead to an insight. Edit: Tl; dr: There is no royal road to Geometry, nor to any advanced Mathematics. Need to explore, struggle until something falls through.

I have got the threshold $k=8086$ for the solution. (It can be worked out by focusing on the discriminant of $x^2 -(k-1)x +2021k=0$) Now, we got to look for an upper limit, if it is possible.

 

[PeroK]

To give some hints for what I did (which is not necessarily the quickest way);

There were two cases: 1) $x = 2021k$ for some integer $k$; and, 2) $x = 43k, y = 47m$ for some $k, m$.

1) leads quickly to a condition on $k - 1$.

For 2) I showed first that $k > 47$ and $m > 43$.

The posts seem to be going in the wrong direction! A bit more help:

The starting equation is$$xy = 2021(x + y + 1) = (43)(47)(x + y + 1)$$1) In the first case we have $x = 2021k$ giving:$$ky = 2021k + y + 1$$Hence$$y = \frac{2021k + 1}{k - 1} = \frac{2021(k-1) + 2022}{k - 1} = 2021 + \frac{2022}{k-1}$$As $y$ is an integer, $k-1$ must be a divisor of $2022$.

The second case is slightly trickier. See the hint.

 

[Prof B]

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

 

[PeroK]

There's a quicker way. Consider $r = x - 2021$ and $s = y - 2021$ and the product $rs$.

 

[anuttarasammyak]

A solution of post #25
Let {P,Q}={1,2021}, {R,S}={1,2022}
we get easily a=d=2022, b=c=2023
x=2022, y=2021*2023＝4088483
I think there is no other solution for these particular P,Q,R,S or as for
$$2021b=2022d-1$$
d=2022 is the only possible number so that RHS is multiple of 2021. But I am not still certain of it.

 

[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
$$l,m,n,r,s=\{0,1\}$$