IMO ShortList 1998, number theory problem 1

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

orl

3647 posts

orl

#1 h Oct 22, 2004, 2:59 PM • 9 Y

Y by Davi-8191, mathematicsy, samrocksnature, Adventure10, megarnie, Mango247, Tastymooncake2, MAS5236, ItsBesi

Determine all pairs $(x,y)$ of positive integers such that $x^{2}y+x+y$ is divisible by $xy^{2}+y+7$ .

Attachments:

This post has been edited 2 times. Last edited by orl, Oct 23, 2004, 12:38 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

orl

3647 posts

orl

#2 h Oct 22, 2004, 2:59 PM • 5 Y

Y by samrocksnature, Adventure10, megarnie, Mango247, Tastymooncake2

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

orl

3647 posts

orl

#3 h Mar 17, 2005, 5:31 PM • 5 Y

Y by samrocksnature, Adventure10, Mango247, MarioLuigi8972, Tastymooncake2

Unchecked solution by nhat:

find all a,b in positive integer such that:
$\frac {a^{2}*b+a+b}{a*b^2+b+7}$ in positive integer
I want the another solution which is different to the solution of IMO
here my solution
for this case a=b we easy to have the solution
if a<b then $a*b*(a-b)+a-7$ <0 (contradition)
hence a>b
let a in $(m*b,(m+1)*b)$ (m i positive integer)
then we easy to prove that : m< $\frac {a^{2}*b+a+b}{a*b^2+b+7}$ <(m+1) (contradition)
so a divisible to b
give $a=k*b$ (k in positive integer)
then
$\frac{k^2*b^3+k*b+b}{k*b^3+b+7}$ = $\frac{b*{k^2*b^2+k+1}}{k*b^3+b+7}$
if $(a,7)=1$ then $(b,k*b^3+b+7)=1$
so that $\frac {k^2*b^2+k+1}{k*b^3+b+7}$ in positive integer
and we esay to prove that k>b
so easy to prove that $k\equiv 0 (mod b)$ so $k=q*b$ (b >= 3and q i positive integer)
then $\frac {k^2*b^2+k+1}{k*b^3+b+7}$ = ${\{q^2*b^4+q*b+1}{q*b^4+b+1}$ = $q$ + $\frac {1-q}{q*b^4+b+1}$ (contradition)
if b divisible to 7 so that $b=7*u$ (u in positive integer)
and easy to prove that u=k (it's similar for this case two)
so we have the solution $a=7*k^2$ , $b=7*k$
my solution hasor hasn't correct

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

paladin8

3237 posts

paladin8

#4 h Mar 6, 2006, 2:24 AM • 5 Y

Y by samrocksnature, Adventure10, Mango247, Tastymooncake2, and 1 other user

The solution given isn't quite correct... there are at least two missing solutions: $(11,1)$ and $(49,1)$ .

$11^2+11+1 = 133$ , which is divisible by $11+1+7 = 19$ .
$49^2+49+1 = 2451$ , which is divisible by $49+1+7 = 57$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

paladin8

3237 posts

paladin8

#5 h Mar 6, 2006, 3:02 AM • 15 Y

Y by acupofmath, DominicanAOPSer, techniqq, Understandingmathematics, myh2910, samrocksnature, Adventure10, Mango247, Tastymooncake2, MAS5236, and 5 other users

My full solution...

If $(xy^2+y+7)|(x^2y+x+y)$ , then there exists a positive integer $k$ such that $k(xy^2+y+7) = x^2y+x+y$ , or

$7k-y = (xy+1)(x-ky)$ .

First we claim that all solutions have both sides nonnegative.

Suppose $7k-y < 0$ . But then we have

$|7k-y| < |y| < |xy+1| < |xy+1||x-ky|$ ,

which is a contradiction. So both sides are nonnegative. Now consider the two cases: (1) both sides are zero, and (2) both sides are positive.

Case 1: Both sides are zero.

Then we have $7k-y = 0 \Rightarrow y = 7k$ . And also $(xy+1)(x-ky) = 0$ , but since $xy+1 > 0$ we know $x-ky = 0 \Rightarrow x = ky = 7k^2$ . So we have all solutions of the form $(x,y) = (7k^2, 7k)$ .

Case 2: Both sides are positive.

Then $x > ky$ so the RHS is positive. So we have $7k > 7k-y = (xy+1)(x-ky) > (xy+1) > y^2k+1$ . Hence $y < 3$ . So we check $y = 1, 2$ .

For $y = 1$ , we have $(x+8)|(x^2+x+1)$ . Since

$\frac{x^2+x+1}{x+8} = x-7+\frac{57}{x+8}$ ,

we have $(x+8)|57 \Rightarrow x = 11, 49$ , giving the solutions $(11,1)$ and $(49,1)$ .

For $y = 2$ , we have $(4x+9)|(2x^2+x+2) \Rightarrow (4x+9)|(4x^2+2x+4)$ . Since

$\frac{4x^2+2x+4}{4x+9} = x-\frac{7x-4}{4x+9}$ ,

so $(4x+9)|(7x-4)$ . But since $2(4x+9) > 7x-4$ , we must have $4x+9 = 7x-4$ , which does not have an integer solution.

Hence our only solutions are $(x,y) = (11,1); (49,1); (7k^2, 7k)$ for positive integers $k$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

abdurashidjon

119 posts

abdurashidjon

#6 h Jun 10, 2006, 11:49 AM • 6 Y

Y by techniqq, samrocksnature, Adventure10, Mango247, Tastymooncake2, and 1 other user

Here my solution I hope there are no mistakes
Let $ab^2+b+7=n$ then $ab^2\cong -b-7 (mod n) \rightarrow a^2b^2\cong -ab-7a (mod n)$
and $a^2b\cong -a-b (mod n) \rightarrow a^2b^2\cong -ab-b^2(mod n)$ from here
$\rightarrow 7a \cong b^2(mod n)$ so $n$ divides both $(ab^2+b+7)$ and $a(b^2-7a)$ so
$(ab^2+b+7)-a(b^2-7a)=b+7+7a^2$ then n divides $b+7+7a^2$ , too.
so $b+7+7a^2\geq ab^2+b+7 \rightarrow 7a \geq b^2$ . Lets share it in two parts
1) $7a-b^2=0 \rightarrow (a,b)=(7k^2,7k)$
2) $7a-b^2>0$
Since we have found that $n|(7a-b^2)$ then $7a-b^2\geq ab^2+b+7$ from here we will have
$(7-b^2)(a+1)\geq b+14$ . $b+14>0$ then $(7-b^2)(a+1)>0$ also $(a+1)>0$ then we have to have $(7-b^2)>0$ here $b$ is positive integer then $b= 1$ or $2$ otherwise inequality will lead to contradiction
If $b=1$ then the problem will become as $\frac{a^2+a+1}{a+8}$
Let $k=a+8$ then $a^2\cong -a-1 (mod k) (i)$
$a\cong -8(mod k)\rightarrow a^2\cong 64 (mod k) (ii)$
from $(i)$ and $(ii)$ we will have $-a-1\cong 64 (mod k)$ and $a+65$ is divisible by $k$ then

$\frac{a+65}{a+8}=\frac{a+8}{a+8} + \frac{57}{a+8}$ so $\frac{57}{a+8} (iii)$ is an integer, too.
so there are just two solutions for $(iii)$ those are $a=11$ or $49$ ( think didisors of $57$ )
If $b=2$ by simliar of previous we will show that there is no solution for $b=2$
So the solutions are $(a,b) = (11,1), (49,1), (7k^2,7k)$
We are done

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

dgreenb801

1896 posts

dgreenb801

#7 h Dec 5, 2009, 4:11 PM • 3 Y

Y by samrocksnature, Adventure10, Mango247

If $x < y$ then it is easy to see there are no solutions.
By the division algorithm, we can let $x = ay + b$ , where $0 \le b < y$ .
Then we have
$a^2y^3 + 2aby^2 + (b^2 + a + 1)y + b$ is divisible by $ay^3 + by^2 + y + 7$
Let the ratio between these two numbers be $a + k$ , then
$a^2y^3 + 2aby^2 + (b^2 + a + 1)y + b = (a + k)(ay^3 + by^2 + y + 7)$
$aby^2 + b^2y + y + b = aky^3 + kby^2 + ky + 7a + 7k$
If $k = 0$ , then $aby^2 + b^2y + y + b = 7a$
Case 1: $b = 0$
Then $y = 7a$ and we have the solution $(7a^2,7a)$
Case 2: $b \ge 0$
Then either $y = 1$ or $2$ . If $y = 1$ , then $b$ can't be more than $7$ , we try out all values from $1$ to $7$ for $b$ . $b = 4$ and $b = 6$ works, this gives the solutions $(11,1)$ and $(49,1)$ . If $y = 2$ then $b = 1$ and we have no solution.
If $k > 0$ , then since $b < y$ , we must have $k = 1$ since if $k > 1$ then the left side is less than the right side (we would have $aky^3 > aby^2 + b$ , $kby^2 > b^2y$ , $ky > y$ ). But if $k = 1$ , then $b = 7a + 7 + (y - b)(ay^2 + by)$ . But $y - b \ge 1$ since $b < y$ , a contradiction.
So $k < 0$ . Then we must have $y = 1$ or the left side would be greater than the right. Since $0 \le b < y$ , we must have $b = 0$ . Thus, $1 = ak + 7a + 8k$ , or $(a + 8)(k + 7) = 57$ . We find $(a,k) = (11, - 3)$ and $(49, - 6)$ work, giving the solutions $(11,1)$ and $(49,1)$ .
So the three solutions are $(7a^2,7a)$ , $(11,1)$ , and $(49,1)$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

KrazyFK

367 posts

KrazyFK

#8 h Apr 5, 2010, 2:05 PM • 6 Y

Y by jam10307, samrocksnature, Adventure10, Mango247, and 2 other users

I thought to myself this morning - "I haven't really tried any IMO problems". I decided to remedy that by solving this one.

First, suppose that $gcd(y,7)=1$ . Then the problem can be restated as follows.

$xy^2+y+7 \, | \, x^2y+x+y$

$\Leftrightarrow xy^2+y+7 \, | \, y(x^2y+x+y)-x(xy^2+y+7)$

$\Leftrightarrow xy^2+y+7 \, | \, y^2-7x$

It is easy to see that since $x,y$ are positive, then $xy^2+y+7 > y^2-7x$ . Then we must have either $y^2=7x$ or $y^2-7x < 0$ . The first case cannot arise since $gcd(7,y)=1$ so we must have $y^2-7x < 0$ .

Then our condition is equivalent to:

$xy^2+y+7 | 7x - y^2$ and we must have $7x - y^2 \ge xy^2 + y + 7$

The second condition implies that $y < 3$ since otherwise we have:

$(7x - y^2) - (xy^2+x+y) \le 7x - 9 -9x -x-3 = -3x - 12 < 0$ .

Hence $y=1$ or $y=2$ .

Case 1: $y=1$ .

Our problem restates as $x+8 | 7x-1$ and so we write:

$\dfrac{7x-1}{x+8} = k$ .

It is clear that $k$ must be positive. We rewrite this as:

$x=\dfrac{8k+1}{7-k}$ and since $x$ is positive, the denominator must be positive and so $k < 7$ . Substituting the values $k=1,2,3,4,5,6$ into this expression in turn shows that only $k=4$ and $k=6$ give integer values for $x$ . These are $x=11$ and $x=49$ .

So our two solutions are $(x,y)=(49,1), (11,1)$ . Checking in the original problem reveals that they work.

Case 2: $y=2$ .

Starting in the same way as above we get $x= \dfrac{9k+4}{7-4k}$ . In a similar way to above we must have $k < 2$ , but substituting $k=1$ does not give integer $x$ . Hence, no solutions in this case.

Now we must consider when $gcd(7,y)=7$ . This case is easy. Let $y=7t$ . Any pair $(x,y)$ which satisfies the problem must also satisfy:

$xy^2+y+7 | y(x^2y+x+y)-x(xy^2+y+7)$

$xy^2+y+7 | y^2-7x$

We cannot have $y=1$ or $y=2$ as above since $gcd(y,7)=7$ . Thus we must have the other possibility, $y^2=7x$ .

Then we have $7x=y^2=49t^2$

So $x=7t^2$

Then our solution is $(x,y)=(7t^2,7t)$ .

Checking this in the original problem:

$\dfrac{x^2y+x+y}{xy^2+y+7}$

$=\dfrac{343t^5+7t^2+7t}{343t^4+7t+7}$

$=t$ which is an integer.

Hence all solutions are given by:

$(x,y) = (11,1), (49,1), (7t^2,7t)$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

JSGandora

4216 posts

JSGandora

#9 h Jan 27, 2012, 1:25 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

KrazyFK wrote:

First, suppose that $gcd(y,7)=1$ . Then the problem can be restated as follows.

$xy^2+y+7 \, | \, x^2y+x+y$

$\Leftrightarrow xy^2+y+7 \, | \, y(x^2y+x+y)-x(xy^2+y+7)$

$\Leftrightarrow xy^2+y+7 \, | \, y^2-7x$

What is the importance of the statement $\gcd(y,7)=1$ ? Can you not perform the same manipulations even if $\gcd(y,7)=7$ ?

Also, my solution involved using the division algorithm to get $\frac{a^2b+a+b}{ab^2+b+7}=\frac{a}{b}+\frac{b^2-7a}{ab^3+b^2+7b}$ , can this only be done if $a\geq b$ ? From there I get $b^2-7a=0\implies b^2=7a\implies a=7k,b=7k^2$ . Why can't I use the division algorithm to get the same result if $b<a$ ?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

GlassBead

1583 posts

GlassBead

#10 h Jan 27, 2012, 1:33 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

The statement of $\gcd(7, y)=1$ is not actually used for the algebraic manipulations, but to set aside a case that forces $y^2-7x$ to be positive (in other words, it's used right after your quote).

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

JSGandora

4216 posts

JSGandora

#11 h Jan 27, 2012, 2:01 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Oh, so from $\frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b}$ , we must split it into cases. So I forgot the case where $b^2-7a<0$ , so $7a-b^2>\frac{ab^{3}+b^{2}+7b}{b}$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

GlassBead

1583 posts

GlassBead

#12 h Jan 27, 2012, 2:10 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Yep.

The $b^2-7a<0$ case leads to the two extra solutions, $(49, 1); (11, 1)$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

JSGandora

4216 posts

JSGandora

#13 h Jan 27, 2012, 2:15 AM • 3 Y

Y by samrocksnature, Adventure10, Mango247

Thanks, but just from the division algorithm, what is the concrete rational that we must have $7a-b^{2}>\frac{ab^{3}+b^{2}+7b}{b}$ (division by $b$ on the RHS)? Is it just that if we factor out the $\frac1b$ on the RHS of $\frac{a^{2}b+a+b}{ab^{2}+b+7}=\frac{a}{b}+\frac{b^{2}-7a}{ab^{3}+b^{2}+7b}$ , we just want $a+\frac{b^2-7a}{ab^2+b+7}$ to be an integer?

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

GlassBead

1583 posts

GlassBead

#14 h Jan 27, 2012, 2:24 AM • 4 Y

Y by JSGandora, samrocksnature, Adventure10, Mango247

That's right. From your division algorithm, if we want the LHS to be an integer, then $b$ has to divide $a+\frac{b^2-7a}{ab^2+b+7}$ , so it must be an integer, and so $ab^2+b+7$ must divide $7a-b^2$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

SoyunSerInerte

6 posts

SoyunSerInerte

#15 h Feb 12, 2014, 10:28 PM • 2 Y

Y by samrocksnature, Adventure10

Define $S=\left \{(a,b)\in \mathbb{Z}^{+}|\displaystyle \frac{a^2b+a+b}{ab^2+b+7}\in \mathbb{Z}^{+}\right \}$

We note that (i)

$ab^2+b+7|a^2b+a+b\Rightarrow ab^2+b+7|a(b(a^2b+a+b)-a(ab^2+b+7))$

then $ab^2+b+7|ab^2-7a^2\Rightarrow ab^2+b+7|ab^2-7a^2-(ab^2+b+7)$

now we have that

$ab^2+b+7|-7a^2-b-7\Rightarrow ab^2+b+7|7a^2+b+7$

note that $ab^2+b+7>0$ and $7a^2+b+7>0$

therefore $ab^2+b+7\leq 7a^2+b+7\Rightarrow b^2\leq 7a$

if $b^2=7a$ then $a=7t^2$ and $b=7t$ for all $t\in \mathbb{Z}^{+}$

therefore $(7t^2,7t)\in S$

now $b^2<7a$

By (i) we have that

$ab^2+b+7|b^2-7a\Rightarrow ab^2+b+7|7a-b^2$ because $b^2<7a$

now because $b^2<7a$ we have $7a-b^2>0$ and note that $ab^2+b+7>0$

therefore $ab^2+b+7\leq 7a-b^2$ that is impossible for $b\ge 3$ therefore $b<3$

We have two cases:

Case 1: $b=1$

we have $a+8|a^2+a+1\Rightarrow a+8|a^2+a+1-a(a+8)+7(a+8)$

therefore $a+8|57=1\cdot 3\cdot 19$ if we prove all the divisors we have that $(49,1)$ and $(11,1)$ are solutions.

therefore $(11,1),(49,1)\in S$

Case 2: $b=2$

we have $4a+9|2a^2+a+2\Rightarrow 4a+9|4(2(2a^2+a+2)-a(4a+9))+7(4a+9)=79$

therefore $4a+9|79$ no solutions here

Finally we have that $S=\{(11,1),(49,1),(7t^2,7t)\}$

Thanks

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

john0512

4187 posts

john0512

#43 h Aug 29, 2023, 12:00 AM

Y by

We claim the answer is $(7n^2,7n),(49,1),(11,1).$ These clearly all work. Note that $x^2y+x+y\equiv 0\pmod{xy^2+y+7}$ $x^2y^2+xy+y^2\equiv 0\pmod{xy^2+y+7}$ $y^2-7x\equiv0\pmod{xy^2+y+7}.$ (at the second step we subtracted $x$ times the modulus.) The point of this is that now everything is linear in $x$ .

Case 1: $y^2-7x>0$ . Then, we have $y^2-7x<y^2\leq xy^2<xy^2+y+7,$ hence there are no solutions as the expression is positive and smaller than the modulus.
f

Case 2: $y^2-7x=0$ . Then, clearly $(x,y)=(7n^2,7n)$ , which clearly works.

Case 3: $y^2-7x<0$ . Then, we have $7x-y^2=c(xy^2+y+7).$ We can rearrange this to $x=\frac{y^2+cy+7c}{7-cy^2}.$
Since $cy^2\leq 6$ , there are not that many cases. If $y=1$ , we have $x=\frac{8c+1}{7-c}.$ For $c=4$ and $c=6$ this gives $x=11$ and $x=49$ , and other values of $c$ up to 6 do not give integer $x$ . If $y=2$ , we have $x=\frac{9c+4}{7-4c},$ here only $c=1$ is allowed which does not give an integer anyway. Finally, $y\geq3$ is not possible, hence done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

huashiliao2020

1292 posts

huashiliao2020

#44 h Sep 4, 2023, 4:14 PM

Y by

We have $0\equiv x^2y+x+y\equiv x^2y^2+xy+y^2-(x(xy^2+y+7))\equiv y^2-7x\pmod{xy^2+y+7}.$ Due to size reasons $y^2-7x\le0$ , if it equals 0 then the solution set is $(7n^2,7n)$ . If it's <0 we must have $xy^2+y+7\le7x-y^2<7x\implies y\in\{1,2\}$ ; easy calculation reduces it into $\boxed{(x,y)\in\{(7n^2,7n),(11,1),(49,1)\}.}\blacksquare$

wait im not sure why but why are my sols so short lol on overleaf typing they take up like 10 lines

This post has been edited 1 time. Last edited by huashiliao2020, Sep 4, 2023, 4:14 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

YaoAOPS

1540 posts

YaoAOPS

#45 h Sep 20, 2023, 7:59 PM

Y by

Note that this implies that $xy^2 + y + 7$ divides $y(x^2y + x + y) - x(xy^2 + y + 7) = y^2 - 7x$ .
This implies that either $y^2 = 7x$ , or that $x \ge 2, y^2 \ge 16$ can not hold.
In the first case, we get the solution set $(x, y) = \left(7k^2, 7k\right)$ which can be seen to hold.
We now bash out the remaining cases. It can be seen that if $x = 1$ , then there are no solutions.
Then, if $y = 1$ , it follows that $x + 8 \mid 1 - 7x$ so $x \in \{11, 49\}$ . We can check that $(x, y) = (49, 1)$ and $(x, y) = (11, 1)$ works.
If $y = 2$ , then $4x + 9 \mid 4 - 7x$ which implies $4x + 9 \mid 3x - 13$ which has no solutions.
If $y = 3$ , then $9x + 10 \mid 9 - 7x$ which has no solutions.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

HamstPan38825

8860 posts

HamstPan38825

#46 h Dec 16, 2023, 7:36 PM

Y by

Rewrite the condition as $xy^2+y+7 \mid (x^2y^2+xy+7x) - (x^2y^2+xy+y^2) = 7x-y^2.$ Now we have a few cases:

If $y^2 > 7x$ , then $y^2 - 7x < y^2 < xy^2+y+7$ , which is obviously impossible. If $y^2 < 7x$ , then $xy^2+(y+y^2) + 7 \leq 7x,$ which implies $y^2 \leq 7$ and $y \in \{1, 2\}$ . This yields the solutions $(49, 1)$ and $(11, 1)$ .

If $y^2 = 7x$ , then the entire curve of solutions $(7n^2, 7n)$ can be checked to work.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

kamatadu

480 posts

kamatadu

#47 h Dec 28, 2023, 7:52 PM

Y by

Bruh, the first time I solved the problem, I solved it for $x^y + x + y \mid xy^2 + y + 7$ ;-; :stretcher:

. Also, these edge cases are so hard for me to find without making sillies.

I claim that the answers are $(x,y)=(11,1)$ , $(49,1)$ and $(7n^2,7n)$ for any $n\in\mathbb N$ .

Firstly, we have $xy^2 + y + 7 \mid x^2y + x + y \mid x^2y^2+xy+y^2\equiv (x^2y^2+xy+y^2)-x(xy^2 + y + 7) = y^2 - 7x$ .

Now if $y^2 > 7x$ , then we get that $xy^2 + y + 7 \le y^2 - 7x$ which gives us $y^2 -y(1+x^2) -(7+7x) \ge 0$ . But then the discriminant must be $\le 0$ , that is $(1+x^2)^2 + 4(7+7x) \le 0$ which clearly has no solution.

Now if $y^2 < 7x$ , then we get that $xy^2 + y + 7 \ge y^2 - 7x$ . Then we get that $x^2y -7x +(y^2+y+7) \le 0$ . This means that the discriminant must be $\ge 0$ , that is $7^2 -4y(y^2+y+7)\ge 0 \implies 4y(y^2+y+7)\le 49$ . This gives only solution $y=1$ since the left side is strictly increasing and it exceeds $49$ for $y=2$ . For $y=1$ , from the problem statement we get that $x+1+7 \mid x^2 + x + 1 \equiv (x^2+x+1)-x(x+8) = -7x+1 \equiv (-7x+1)+7(x+8) = 57$ . This gives us that $x\in \left\{11,49\right\}$ each of which are solutions.

Now for the other case when $y^2 = 7x$ , then clearly $7\mid y$ and so $49n^2 = 7x \implies x = 7n^2$ which is another solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

shendrew7

795 posts

shendrew7

#48 h Mar 11, 2024, 8:22 PM

Y by

Notice the LHs also divides
$y(x^2y+x+y)-x(xy^2+y+7) = y^2-7x.$
If $y^2-7x=0$ , we have the solutions $\boxed{(7k^2,7k)}$ . Otherwise, we notice
$|xy^2+y+7| > |y^2-7x|,$ implying there are no solutions, unless $y=1,2$ where we get the pairs $\boxed{(11,1),(49,1)}$ . $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

MagicalToaster53

159 posts

MagicalToaster53

#49 h Jun 2, 2024, 8:58 PM

Y by

I claim that all solutions are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t), \}}$ , for $t \in \mathbb{Z}^+$ .

Observe that $ab^2 + b + 7 \mid a^2b + a + b \implies ab^2 + b + 7 \mid b(a^2b + a + b) - a(ab^2 + b + 7) = b^2 - 7a.$ We now split into two separate cases:

Case 1: $(b^2 - 7a = 0).$ Then $b^2 = 7a$ , so that $b \equiv 0 \pmod 7$ , which in turn gives us $b = 7t$ , for some $t \in \mathbb{Z}^+$ . Then we find the corresponding $a = 7t^2$ . $\square$

Case 2: $(b^2 - 7a < 0).$ First observe that $b^2 - 7a \ngeq 0$ , else $b^2 - 7a \geq ab^2 + b + 7 \implies b^2 \geq a(b^2 + 7) + b + 7$ , which is a clear contradiction for $a, b > 0$ . Hence $b^2 - 7a < 0$ , so that we obtain $ab^2 + b + 7 \leq 7a - b^2 \implies b^2(a + 1) + b + 7 \leq 7a \implies b^2 < 7.$ Hence we split into cases for $b = 1, 2$ :

Subcase 2.1: $(b = 1).$ Then $\frac{7a - 1}{a + 8} \in \mathbb{Z} \implies 7 - \frac{57}{a + 8} \implies a + 8 = 1, 3, 19, 57 \implies \boxed{a = 11, 49}. \bigstar$
Subcase 2.2: $(b = 2).$ Then $\frac{7a - 4}{4a + 9} \in \mathbb{Z} \implies 1 + \frac{3a - 13}{4a + 9} \implies a \leq -22,$ which is impossible. Therefore no solution exists in this subcase. $\bigstar$

The only solutions, therefore, are $\boxed{(a, b) \in \{(11, 1), (49, 1), (7t^2, 7t))\}}$ , for arbitrary $t \in \mathbb{Z}^+$ , as claimed. $\blacksquare$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Aryan27

40 posts

Aryan27

#50 h Jul 9, 2024, 4:46 PM • 1 Y

Y by GeoKing

The solutions are $(x,y) = (11,1)$ , $(49,1)$ , $(7k^2,7k)$ for all $k\in\mathbb N$ .

Note that, we are given that:
$\begin{align*} xy^2+ y+ 7\mid x^2y + x + y \implies xy^2+ y+7\mid y(x^2y + x + y) - x(xy^2+y+7)= y^2-7x \end{align*}$
Now we divide into cases based on the sign of $y^2-7x$

When $y^2-7x> 0$ .
The divisibility condition implies that $y^2-7x\geq xy^2+ y+ 7$
Clearly, $0<y^2-7x<xy^2+ y+ 7$ , contradicting the divisibilty condition.

When $y^2-7x=0$ .
in this case we get ,
$y^2=7x$ , let $y = 7k$ , so $x = 7k^2$ .
Plugging this back in to the original equation reads:
$\begin{align*} 343k^4 + 7k + 7 \mid 343k^5 + 7k^2 + 7k \end{align*}$ which is always valid, hence these are always solutions.

When $y^2-7x<0$ .
We get:
$\begin{align*} |y^2-7x|\geq xy^2+ y+ 7 \implies 7x-y^2\geq xy^2+y+7 \iff x(y^2-7)+y^2+y+7\le 0 \iff y \in \{1,2\}. \end{align*}$
When $y=1$ we get:
$\begin{align*} x+8 \mid 7x-1 \iff x+8 \mid 57 \end{align*}$ This gives $x=11$ and $x=49$ .

When $y=2$
$\begin{align*} 4x+9 \mid 7x-4\iff 4x+9 \mid 79 \end{align*}$ which gives no solutions.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

RedFireTruck

4223 posts

RedFireTruck

#51 h Aug 8, 2024, 6:05 AM

Y by

Assume that $\gcd(y, 7)=1$ . Then, $\gcd(xy^2+y+7,x^2y+x+y)=\gcd(xy^2+y+7, y^2-7x)=\gcd(7x^2+y+7, 7x-y^2).$
We want this to equal $xy^{2}+y+7$ , so $7x^2+y+7\ge xy^2+y+7$ so $7x\ge y^2$ .

We also want $7x-y^2\ge xy^2+y+7$ or $7(x-1)\ge (x+1)y(y+1)$ . This means that $y=1$ or $y=2$ . When $y=1$ , we get $(x+8)|(x^2+x+1)$ so $(x+8)|57$ so $x=11$ or $x=49$ .

When $y=2$ , we get $(4x+9)|(2x^2+x+2)$ so $(4x+9)|(4x^2+2x+4)$ so $(4x+9)|(x+22)$ so there are no solutions to $x$ .

Now assume that $y=7b$ . Then, $7|(xy^{2}+y+7)|(x^{2}y+x+y)$ so $x=7a$ . Plugging this in means $(49ab^2+b+1)|(49a^2b+a+b)$ . Note that $\gcd(49ab^2+b+1, 49a^2b+a+b)=\gcd(49ab^2+b+1, b^2-a).$
Note that $|b^2-a|< 49ab^2+b+1$ so $a=b^2$ . Plugging $a=b^2$ back in, we get $(49b^4+b+1)|(49b^5+b^2+b)$ , which is always true.

Therefore, the solutions are $(11,1)$ , $(49,1)$ , and $(7k^2,7k)$ for all positive integer $k$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ezpotd

1263 posts

ezpotd

#52 h Aug 21, 2024, 3:54 AM

Y by

Observe that we then have $xy^2 + y + 7 \mid x^2y^2 + xy + y^2$ , so $xy^2 + y + 7 \mid y^2 - 7x$ . We can then divide into cases, if $y^2 > 7x$ , then clearly we have no solutions by size. If $y^2 = 7x$ , then write $y = 7k, x = 7k^2$ , we write $x^2y + x + y = 343k^5 + 7k^2 + 7k, xy^2 + y + 7 = 343k^4 + 7k + 7$ , we can see all solutions of this form work. If $y^2 < 7x$ , by size we still require $7x - y^2 \ge xy^2 + y + 7$ , or equivalently $y^2 < 7$ . We then check $y = 2$ , we require $4x + 9 \mid 7x - 4$ , equivalently $4x + 9 \mid 28x - 16$ , equivalently $4x +9 \mid 79$ , so there are no solutions for $x$ by divisor analysis. We now check $y = 1$ , we require $x + 8\mid 7x - 1$ , so we have $x + 8 \mid 57$ , so we have $x = 11, 49$ . We check $x = 11$ gives $x^2y + x + y = 133$ , $xy^2 + y + 7 = 19$ , so this pair works. We then check $x = 49$ , we get $x^2y + x + y = 2451, xy^2 + y + 7 = 57$ , so this pair works as well. The answers are then $(7k^2, 7k), (49,1), (11,1)$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Flint_Steel

38 posts

Flint_Steel

#53 h Sep 8, 2024, 9:55 AM

Y by

\Rightarrow supremacy :wacko:

$ab^2+b+7|a^{2}b+a+b \Rightarrow b(ab+1)+7|ab(ab+1)+b^2 \Rightarrow ab^2+b+7|b^2-7a$
$ab^2+b+7|ab^2-7a^2 \Rightarrow ab^2+b+7|7a^2+b+7$ . Since both sides are positive: $ab^2 \leq 7a^2 \Rightarrow b^2\leq 7a$ .
So there is two cases to consider.
First case $b^2=7a$ : if we set $b=7k$ , then $a=7k^2$ , We can easily check that it is a solution with $k$ being a positive integer.
Second case $b^2<7a$ : Then from earlier, $ab^2+b+7<7a-b^2 \Rightarrow b^2+b+7<a(7-b^2)$ LHS is positive so RHS should follow. Meaning
$7>b^2$ . Then we can manually check and see that $(a,b)=(49,1); (11,1)$ is a solution.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

math004

23 posts

math004

#54 h Jan 26, 2025, 8:04 AM

Y by

$xy^2+y+7 \mid y(x^2y+x+y)-x(xy^2+y+7)=y^2-7x$ which implies that $|y^2-7x| \geq xy^2+y+7$ which gives $x=1$ or $y^2\leq 7.$ or $y^2=7x\implies (x,y)=(7t^2,7t)$ which Convsersely always works. The edge cases give $(1,49)$ and $(1,11)$ as solutions.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

pie854

243 posts

pie854

#55 h Feb 13, 2025, 9:36 AM

Y by

A bit of long division leads us to $xy^2+y+7 \mid y^2-7x$ . Clearly $y^2-7x>0$ isn't possible due to size. If $y^2-7x<0$ then $7x>7x-y^2>xy^2+y+7>xy^2 \implies y=1,2.$ After checking we find the solution $(11,1),(49,1)$ . If $y^2=7x$ then $(x,y)=(7k^2,7k)$ for some $k$ , which works.

This post has been edited 1 time. Last edited by pie854, Feb 13, 2025, 9:36 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

Ilikeminecraft

616 posts

Ilikeminecraft

#56 h Apr 25, 2025, 4:14 PM

Y by

We first consider when $(y, 7) = 1.$ Multiply the RHS by $y$ to get $x^2y^2 + xy + y^2,$ and then subtracting $x$ times the LHS, we get $y^2 - 7x.$ Thus, we have that $xy^2 + y + 7 \mid y^2 - 7x.$

If $y^2 > 7x,$ we have that $y^2 - 7x \geq xy^2 + y + 7,$ but this is absurd.

If $7x > y^2,$ we have that $7x - y^2\geq xy^2 + y + 7.$ Thus, $y = 2, 1.$ If $y = 1,$ we have $x + 8\mid7x - 1.$ Clearly, the solutions are $(11, 1), (49, 1).$ If $y = 2,$ we have that $4x + 9 \mid 7x - 4.$ Multiplying by $4,$ we have that $4x + 9 \mid -79,$ which has no solutions.

Now, we consider $(y, 7) = 7.$ Thus, we have that $y = 7k.$ We have $343xk^2 + 7k + 7 \mid 7x^2k + x + 7k.$ We clearly have that $x = 7m$ for some $m\in\mathbb N.$ Thus, $343mk^2 + k + 1 \mid 343 m^2k + m + k.$ The LHS is very clearly relatively prime to $k,$ so we multiply the RHS by $k$ and then apply Euclids, we have that $343mk^2 + k + 1 \mid k^2-m.$ If it is non-zero, this is clearly absurd. Thus, $m = k^2.$ We get the curve $(7k^2, 7k).$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

reni_wee

45 posts

reni_wee

#57 h Yesterday at 4:00 PM

Y by

$\begin{align*} xy^2 + y + 7 & \mid x^2y + x + y \\ \implies xy^2 + y + 7 & \mid x(x^2y + x + y) - y(xy^2 + y + 7) \\ \implies xy^2 + y + 7 & \mid y^2 -7x \\ \end{align*}$ We now proceed to solve this problem using 3 cases.

Case i. $y^2 - 7x > 0$
$\implies y^2 -7x \geq xy^2 + y + 7$ As $x$ and $y$ are positive integers,
$y^2 -7x < y^2 < xy^2 + y + 7$
Hence a contradiction.

Case ii. $y^2 - 7x = 0$
$\implies y^2 = 7x$ $\therefore (x,y) = (7k^2,7k)$ ; $k \in \mathbb{Z^+}$ work.

Case iii. $y^2 - 7x < 0$
$\implies 7x -y^2 \geq xy^2 + y + 7$ For $y^2 > 7$ we have,
$7x - y^2 < 7x < xy^2 + y + 7$ which is a contradiciton.
Hence $y^2 \leq 7 \implies y \leq 2$ . Therefore we only need to consider the cases where $y=1$ and $y = 2$

When $y = 1$ ,
$\begin{align*} x+8 & \mid 7x -1 \\ \implies x+8 & \mid 7(x +8) - (7x-1) \\ \implies x+8 & \mid 57 \end{align*}$ Hence, $(x,y) = (11,1), (49,1)$ works.

When $y=2$ ,
$\begin{align*} 4x+9 & \mid 7x -4 \\ \implies 4x+9 & \mid 7(4x +9) - 4(7x-4) \\ \implies 4x+9 & \mid 79 \end{align*}$ $\implies x = 17.5 \not \in \mathbb{Z^+}$ .

Therefore the only solutions are $(11,1), (49,1)$ and $(7k^2, 7k)$ ; $k \in \mathbb{Z^+}$

This post has been edited 2 times. Last edited by reni_wee, Yesterday at 5:19 PM

Z K Y