IMO Problem 2

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iandrei

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iandrei

#1 h Jul 14, 2003, 5:06 PM • 10 Y

Y by Davi-8191, biomathematics, VladimirArauzo050, Adventure10, megarnie, Mango247, and 4 other users

Determine all pairs of positive integers $(a,b)$ such that $\dfrac{a^2}{2ab^2-b^3+1}$ is a positive integer.

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Anonymous

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Anonymous

#2 h Jul 15, 2003, 3:16 PM • 7 Y

Y by IFA, Adventure10, megarnie, Mango247, and 3 other users

First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1
and so, 1<=b<=2a.
Now, we'll prove that the pairs that solve this problem are
(i) (a,b)=(n,1),
(ii) (a,b)=(n,2n), and
(iii) (a,b)=(8n^4-n,2n)
where n is a positive

Indeed, if b=1, we get the pairs in (i), while if b=2a we get the
pairs in (ii). Now assume that 1<b<2a and set r=a^2/( 2ab^2-b^3+1)
Then the quadratic equation

(*) x^2-2xrb^2+r(b^3-1)=0

has two solution that are positive integers, since one of them is a and their product is r(b^3-1)>0. Evaluating the discriminant D we have
D=4(r^2b^4-rb^3+r). Thus r^2b^4-rb^3+r=m^2 for some natural
number and the above equation has solution rb^2+m and rb^2-m. Now
note that

m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,

with rb^2-b>0. Hence, b>rb^2-m.
However, since a^2> b^2(2a-b>=b^2, we have a>b. Therefore,
a=rb^2+m > b and the other (also positve) root, say c, is c=rb^2-m < b.
Furthermore, if 2c>b, then we have

r=c^2/(b^2(2c-b)+1 <= c^2/b^2 <1,

a contradiction. Thus b=2c and r=c^2. Since ac=r(b^3-1), we get
a=c(b^3-1)=8c^4-c. It is easy to verify that if c>=1, then the integers
b=2c and a=c(b^3-1) solve the problem, since

a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2.

The proof is complete.

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liyi

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liyi

#3 h Jul 18, 2003, 8:09 AM • 4 Y

Y by Adventure10, DoThinh2001, Mango247, and 1 other user

if b = 1, a is an even number.
so case(i) may be (2k,1)

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Anonymous

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Anonymous

#4 h Jul 18, 2003, 9:22 AM • 3 Y

Y by Adventure10, Mango247, and 1 other user

why the denominator has to be positive? couldnt it be negative?

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Anonymous

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Anonymous

#5 h Jul 18, 2003, 1:27 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Oh, yes...when b=1, then a must be even so case (i) becomes (n,1) with n even. Sorry about that!

The denominator must be positive since the numerator is positive and the ratio must be a positive integer.

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Anonymous

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Anonymous

#6 h Jul 22, 2003, 3:10 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

sorry I didnt know that the ratio has to be positie integer. all i know was it has to be integer...

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paganinio

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paganinio

#7 h Dec 23, 2004, 12:58 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

Be careful next time.
a <sup>2</sup> >=0 so DENOminator >=0

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Davron

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Davron

#8 h May 26, 2006, 1:08 AM • 6 Y

Y by Mediocrity, Adventure10, megarnie, Gaunter_O_Dim_of_math, Mango247, and 1 other user

Who can help me with Dmitri Shiryayevs elegant solution :
[russian team member imo 2003]

Let $n=2ab^2-b^3+1$ than we have $2ab^2 \equiv b^3-1 (modn)$ , also we have
$2a^2b^2 \equiv ab^3-a (modn)$ but $a^2 \equiv 0 (modn)$ . Thus we have $ab^3-a \equiv 0 (modn)$ ,
$2ab^3-2a \equiv 0 (modn)$ also $2ab^3 -b^4+b \equiv 0 (modn)$ since n divides n.
So we have $2a \equiv b^4 -b (modn)$ .
That gives: $m=b^4-b-2a \equiv 0 (modn)$ .
And, $n=(2a-b)b^2+1>0$ since $\frac{a^2}{2ab^2-b^3+1}$ is positive and $a^2$ postitive, too. So that means denominator is also positive. By the way we have $2a-b \geq 0$
1. $2a=b$ gives $(a,b)=(k,2k)$
2. $2a-b>0$ ... $2a-b \geq 1$ . It follows that $a^2$ divisible by $n$ .
$a^2 \geq n = (2a-b)b^2+1$ ... $a^2 \geq b^2+1$ than we have $a>b^2$ $(*)$
And $n=(2a-b)b^2+1>(2b^2-b)b^2+1=2b^4-b^3+1>b^4>b^4-b-2a=m$
From here we have $0 \geq m$ .
a-) $b^4-b-2a=0$ gives series $(a,b)=(8k^4-k,2k)$
b-)If $b^4-b-2a<0$ so we have $-(b^4-b-2a) \geq n=2ab^2-b^3+1$ than
$2a(1-b^2) \geq b^4-b-b+1=(b-1)(b-1)^3$
So $b=1$ which gives $(2k,1)$ .

So we are done.

Please explain me : $(*)$

davron

This post has been edited 1 time. Last edited by Davron, Jun 7, 2006, 3:28 AM

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abdurashidjon

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abdurashidjon

#9 h May 27, 2006, 12:27 PM • 7 Y

Y by karitoshi, Adventure10, Mango247, and 4 other users

Achilleas Sinefakopoulos wrote:

First note that we must have 1=< 2ab^2-b^3+1=b^2(2a-b)+1
and so, 1<=b<=2a.
Now, we'll prove that the pairs that solve this problem are
(i) $(a,b)=(n,1),$
(ii) $(a,b)=(n,2n)$ , and
(iii) $(a,b)=(8n^4-n,2n)$
where $n$ is a positive

Indeed, if $b=1$ , we get the pairs in (i), while if b=2a we get the
pairs in (ii). Now assume that $1<b<2a$ and set $r=a^2/( 2ab^2-b^3+1)$
Then the quadratic equation

(*) $x^2-2xrb^2+r(b^3-1)=0$

has two solution that are positive integers, since one of them is a and their product is $r(b^3-1)>0$ . Evaluating the discriminant D we have
$D=4(r^2b^4-rb^3+r)$ . Thus $r^2b^4-rb^3+r=m^2$ for some natural
number and the above equation has solution $rb^2+m$ and $rb^2-m$ . Now
note that

$m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$

with $rb^2-b>0$ . Hence, $b>rb^2-m$ .
However, since $a^2> b^2(2a-b\geq b^2$ , we have $a>b$ . Therefore,
$a=rb^2+m > b$ and the other (also positve) root, say c, is $c=rb^2-m < b$ .
Furthermore, if 2c>b, then we have

$r=c^2/(b^2(2c-b)+1 \leq c^2/b^2 <1$ ,

a contradiction. Thus $b=2c$ and $r=c^2$ . Since $ac=r(b^3-1)$ , we get
$a=c(b^3-1)=8c^4-c$ . It is easy to verify that if $c\geq 1$ , then the integers
b=2c and a=c(b^3-1) solve the problem, since

$a^2/(2ab^2-b^3+1) = c^2(b^3-1)^2 / [(b^3-1)(2cb^2-1)]=c^2$ .

The proof is complete.

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bilarev

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bilarev

#10 h May 30, 2006, 10:11 AM • 3 Y

Y by Bunrong123, Adventure10, Mango247

Davron wrote:

$a^2 \geq b^2+1$ than we have $a>b^2$

I cant understand this...

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Davron

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Davron

#11 h May 30, 2006, 11:39 AM • 2 Y

Y by Adventure10, Mango247

what do you think about that,
but it might be true since i read that from KVANT magazine
[ i hope that someone will help]

davron

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Davron

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Davron

#12 h Jun 7, 2006, 4:05 AM • 1 Y

Y by Adventure10

bilarev wrote:

Davron wrote:

$a^2 \geq b^2+1$ than we have $a>b^2$

I cant understand this...

ok now i got this point

bilarev look it is really nice

we know that $a^2\geq 2ab^2-b^3+1=b^2(2a-b)+1>b^2a$ thus we have $a^2>b^2a$ implies to $a>b^2$
nice right ?

davron

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yookon88

65 posts

yookon88

#13 h Jun 7, 2006, 2:50 PM • 2 Y

Y by Adventure10, Mango247

Davron wrote:

bilarev wrote:

Davron wrote:

$a^2 \geq b^2+1$ than we have $a>b^2$

I cant understand this...

ok now i got this point

bilarev look it is really nice

we know that $a^2\geq 2ab^2-b^3+1=b^2(2a-b)+1>b^2a$ thus we have $a^2>b^2a$ implies to $a>b^2$
nice right ?

davron

I don't understand how from $a^2\geq b^2(2a-b)+1$ you get $a^2>b^2a$

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Davron

484 posts

Davron

#14 h Jun 7, 2006, 4:35 PM • 2 Y

Y by Adventure10, Mango247

ok i will explain this more clearly :

we know that $2ab^2-b^3+1$ is positive since $a^2\geq 0$
Also we know that $a^2 \geq b^2(2b-b)+1>b^2+1$ from here we can easily get that $a>b$ .
ok ?
Then we again rewwrite the inequality $a^2 \geq b^2(2a-b)+1=b^2[a+(a-b)]+1>b^2a+1$ since $a>b$
Thus we can easily see that $a^2>b^2a$ which leasds to the desired $a>b^2$

[please if you understand this let me know

]

davron

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yookon88

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yookon88

#15 h Jun 7, 2006, 5:49 PM • 1 Y

Y by Adventure10

i understand... thank you

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mst.4921

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mst.4921

#35 h Mar 2, 2021, 5:26 PM

Y by

jgnr wrote:

Click to reveal hidden text

that's wonderful

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jeteagle

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jeteagle

#36 h Jul 25, 2021, 2:33 AM

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The solutions are $(x, 2x), (8x^4 -x, 2x), (2x, 1)$ for all positive integers $x$ , which can be easily verified to work.

Let $a^2 = k(2ab^2-b^3+1)$ for some positive integer $k$ . We can solve to get: $a^2 = k(2ab^2-b^3+1) \implies a^2-a(2kb^2)+k(b^3-1) = 0 \implies a = \frac{2kb^2\pm\sqrt{(2kb^2)^2-4k(b^3-1)}}{2},$ so the discriminant is $4k^2b^4-4kb^3+4k.$ For $a$ to be an integer this must be a square, so set this equal to $n^2$ for some positive integer $n$ . We have that $4k^2b^4-4kb^3+4k = (2kb^2-b)^2-(b^2-4k) = n^2.$ We have three cases: $b^2-4k > 0$ , $b^2-4k < 0$ , or $b^2-4k = 0$ .

Case 1: $b^2-4k > 0.$

We have $b^2-4k = (2kb^2-b)^2-n^2 \ge (2kb^2-b)^2-(2kb^2-b-1)^2 = 4kb^2-2b-1 > b^2-4k,$ a contradiction.

Case 2: $b^2-4k < 0$ .

We have $4k-b^2 = n^2-(2kb^2-b)^2 \ge (2kb^2-b+1)^2-(2kb^2-b)^2 = 4kb^2-2b+1.$ Notice that if $b \ge 2$ , then $4kb^2-2b+1 > 4k-b^2$ , a contradiction. Only if $b = 1$ , then $4kb^2-2b+1 \le 4k-b^2$ , and we actually achieve equality when $n = 2kb^2-b+1$ . Plugging in $b = 1$ gives $a = 2x$ for all positive integers $x$ , and we have our first set of solutions: $(2x, 1)$ .

Case 3: $b^2-4k = 0$ .

This means $n = 2kb^2-b$ and $4k = b^2$ , so solving for $a$ , we get $a = \frac{2kb^2 \pm (2kb^2-b)}{2},$ so $a = \frac{b}{2}$ , so we get our second set of solutions $(x, 2x)$ , or $a = \frac{4kb^2-b}{2} = \frac{b^4-b}{2}$ for even $b$ . Substituting $b = 2c$ gives $a = 8c^4-c$ for all positive integers $c$ , which gives our final set of solutions $(8x^4-x, 2x)$ . $\blacksquare$

Remarks: Nice discriminant bounding problem. This problem was quite straight forward and wasn't as contrived as I thought it would be.

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awesomeming327.

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awesomeming327.

#37 h Mar 21, 2022, 12:32 AM

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When in doubt, do some bounds.

Let $a^2=k(2ab^2)-b^3k+k$ then $a^2-(2b^2k)a+b^3k-k,$ so by the quadratic formula $a=b^2k\pm \sqrt{b^4k^2-b^3k+k}$

For $a$ to be an integer, $b^4k^2-b^3k+k$ must be a perfect square. If $b=1$ then clearly all even $a$ work, so let $b>1.$

Let $r=b^2k-\frac{b}{2}+c$ then $r^2=b^4k^2-b^3k+\frac{b^2}{4}+2b^2kc-bc+c^2.$ If $b^4k^2-b^3k+k=r^2$ then $\frac{b^2}{4}+2b^2kc+c^2=bc+k.$

If $c$ is positive, then $b^2kc\ge bc$ and $b^2kc\ge k$ so $\frac{b^2}{k}+c^2\le 0$ implying $c=0$ a contradiction.

If $c$ is negative, then let $c=-c_0$ note that $\frac{b^2}{4}-bc+c^2=k(1-2b^2c)$ so $1+2b^2c_0\le \frac{b^2}{4}+bc_0+c_0^2.$ Note that for $c_0\ge 1$ we have $2c_0< \frac{1}{4}$ while $1\le bc_0+c_0^2$ so we have another contradiction. Thus, $c=0.$ This implies that $k=\frac{b^2}{4}.$

We can plug in for $a.$ Thus, we have the following solutions:

$(2k,1),(k,2k),(8k^4-k,2k)$ for positive integer $k$ which all work.

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ZETA_in_olympiad

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ZETA_in_olympiad

#38 h Apr 13, 2022, 8:57 AM

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#2 in $\LaTeX$ .

First note that we must have $1\leq 2ab^2-b^3+1=b^2(2a-b)+1$ And so, $1\leq b \leq 2a.$ Now, we'll prove that the pairs that solve this problem are $(a,b)=(n,1),(n,2n),(8n^4-n,2n),$ where $n$ is positive.

Indeed, if $b=1,$ we get the first pair, while if $b=2a$ we get the second
pair.

Now assume that $1<b<2a$ and set $r=\frac{a^2}{ 2ab^2-b^3+1}.$
Then the quadratic equation

$x^2-2xrb^2+r(b^3-1)=0$
has two solution that are positive integers, since one of them is $a$ and their product is $r(b^3-1)>0.$ Evaluating the discriminant $D$ we have
$D=4(r^2b^4-rb^3+r).$ Thus $r^2b^4-rb^3+r=m^2$ for some natural
number and the above equation has solution $rb^2+m$ and $rb^2-m.$ Now note that

$m^2 = r^2b^4-rb^3+r > r^2b^4-2rb^3+b^2=(rb^2-b)^2,$
with $rb^2-b>0.$ Hence, $b>rb^2-m.$
However, since $a^2> b^2(2a-b) \geq b^2$ , we have $a>b.$ Therefore,
$a=rb^2+m > b$ and the other (also positive) root, say $c,$ is $c=rb^2-m < b.$
Furthermore, if $2c>b$ , then we have

$r=c^2/(b^2(2c-b)+1 \leq c^2/b^2 <1,$
a contradiction. Thus $b=2c$ and $r=c^2$ . Since $ac=r(b^3-1)$ , we get
$a=c(b^3-1)=8c^4-c$ . It is easy to verify that if $c\geq 1$ , then the integers
$b=2c$ and $a=c(b^3-1)$ solve the problem, since

$\frac{a^2}{2ab^2-b^3+1}= \frac{c^2(b^3-1)^2}{(b^3-1)(2cb^2-1)}=c^2.$
The proof is complete.

This post has been edited 2 times. Last edited by ZETA_in_olympiad, Apr 13, 2022, 9:02 AM

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ZETA_in_olympiad

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ZETA_in_olympiad

#45 h Apr 14, 2022, 1:03 PM

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Let $(a,b)$ satisfy $\frac{a^2}{2ab^2-b^3+1}=m, ~~~ m\in \mathbb{N}.$ It can be simplified to $(2(a-b^2m))^2=(2b^2m-b)^2+4m-b^2. ~~~~~(1)$

Case 1: $4m-b^2>0.$

By (1) we have $\begin{align*}4m-b^2=(2(a-b^2m))^2-(2b^2m-b)^2 \\ \geq (2b^2m-b+1)^2-(2b^2m-b)^2=2(2b^2m-b)+1 \end{align*}$
This is true because we can set $M=2(a-b^2m), N=2b^2m-b, ~~~ M,N \in \mathbb{N}.$ Due to $M^2-N^2>0$ and $N>0,$ $\begin{align*} |M| \geq N+1 \\ \implies M^2-N^2 \geq (N+1)^2 -N^2=2N+1 \\ \implies 2(2b^2m-b)+1+b^2-4m \leq 0 \\ \implies 4m(b^2-1)+(b-1)^2 \leq 0. \end{align*}$ So it must be that $b=1,$ giving $\boxed{(a,b)=(2m,1)}$ as a satisfying pair.

Case 2: $4m-b^2<0.$

By (1) we have, $\begin{align*} 4m-b^2&= (2(a-b^2m))^2-(2b^2m-b)^2 \\ \leq (2b^2m-b-1)^2 -(2b^2m-b)^2=-2(2b^2m-b)+1 \\ \implies (4m-1)b^2-2b+(4m-1)\leq 0~~~~~ (2) \\ \implies D=4(1-(4m-1)^2)< 0, \end{align*}$ where $D$ is the discriminant of the quadratic polynomial of $b$ .

Therefore because $4m-1>0,$ (2) is not true. And thus there is no pair $(a,b)$ satisfying the property in this case.

Case 3: $4m-b^2=0.$

So, $(2(a-b^2m))^2=(2b^2m-b)^2 \implies 2a-2b^2m=\pm (2b^2k-b).$
Subcase 1:

$2a-2b^2m=-2b^2m+b$ implies that $b=2a.$ So we get the pair $\boxed{(a,b)=(k,2k)}$ where $k$ is any natural number, as a satisfying pair.

Subcase 2:

$2a-2b^2m=2b^2m-b \implies 4b^2m-b=2a.~~~~~(3)$ However $4m-b^2=0$ implies $b$ is even. So there exists a $c$ such that $b=2c$ and $m=c^2.$ By (3) we get $a=8c^4-c.$

So another pair is $\boxed{(a,b)=(8c^4-c, 2c)}$ and we are done satisfying all cases.

This post has been edited 2 times. Last edited by ZETA_in_olympiad, Apr 14, 2022, 7:00 PM

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megarnie

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megarnie

#46 h Jul 7, 2022, 1:21 AM

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Solved with pi271828 a while back.

The answer is $\boxed{(2n,1), (n,2n), (8n^4-n,2n)}$ for any positive integer $n$ . These work. We now prove they are the only solution.

Let $\frac{a^2}{2ab^2-b^3+1}=k$ .

Clearly if $b=1$ , then $k=\frac{a}{2}$ , so $a$ must be even. This falls in the solution set ( $(2n,1)$ ).

Then we have $a^2=2ab^2k - kb^3 + k = a^2 - 2kb^2 \cdot a + (kb^3 - k) = 0$
Claim: Either $b=2a$ or $a>b$ .
Proof: Suppose $a\le b$ .

Note that $a^2\ge 2ab^2 - b^3 + 1 > 0$ .

So $a^2 > b^2 (2a-b)>-1$ .

This implies $2a-b\ge 0$ .

If $2a-b>0$ , then we must have $a^2>b^2$ , contradiction. So $b=2a$ . $\square$

Now, consider the quadratic $X^2-2kb^2 X + (kb^3 - k) = 0$
Let the two roots of this equation be $a_0$ and $a_1$ , with $a_0\ge a_1$ .
We have $\begin{align*} a_0+a_1 = 2kb^2 \\ a_0\cdot a_1 = kb^3 - k = k(b^3-1) \\ \end{align*}$
If $a_0= a_1$ , then $a_0=a_1=kb^2$ , so $(kb^2)^2 = k(b^3-1)\implies kb^4 = b^3 - 1,$ not true.

So we have $a_0>a_1$ .

This implies $a_0>kb^2$ .

If $b=2a_0$ , then $b\ne 2a_1$ , which implies $a_1>b$ , however $b=2a_0>2a_1$ . So $b\ne 2a_0$ .

Case 1: $b=2a_1$ .
Then we have $a_0 = 2kb^2 - \frac{b}{2}$ , so $\left(2kb^2 - \frac{b}{2}\right)\cdot \frac{b}{2} = kb^3 - k$ , which implies $kb^3 - \frac{b^2}{4} = kb^3 - k$ ., so $k=\frac{b^2}{4}$ .

So $a_0 = \frac{b^4-b}{2}$ .

Setting $b=2a_1$ gives the solutions $(a_0, b) = (8a_1^4 - a_1, 2a_1)$ and $(a_1,b) = (a_1, 2a_1)$ , both of which fall into the solution set described above ( $(8n^4-n,2n)$ and $(n,2n)$ ).

Case 2: $b\ne 2a_1$ .
In this case, $a_0>a_1>b$ .

Since $a_0>kb^2$ , we have $a_0\cdot a_1 > kb^2 \cdot b = kb^3$ , which implies $a_0\cdot a_1\ne k(b^3-1)$ , contradiction.

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HamstPan38825

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HamstPan38825

#47 h Apr 23, 2023, 2:03 PM

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This is a pretty nice Vieta jumping problem.

The answers are $(2n, 1), (n, 2n), (8n^2-4, 2n)$ for all positive integers $k$ . They obviously work. Assume $b>1$ in what follows.

Fix $k$ to be the value of the expression, and assume it is an integer. Then, the condition is equivalent to $a^2-2ab^2k+b^3k - k = 0.$ In other words, if $(a, b)$ is a solution, then so is $\left(\frac{b^3k-k}a, b\right).$

Claim. We will have $b^3k-k < a^2$ unless $2a=b$ or $a \leq b$ .

Proof. Note that $b^3k-k < b^3k <\frac{a^2b}{2a-b}$ , so the condition simplifies to $a > b$ . $\blacksquare$

On the other hand, if $a\leq b$ in the original equation, then we have $a^2 > b^2(2a-b) > -1$ , which implies that $b=2a$ as otherwise $a^2>b^2$ . As a result, this means that the minimal solution $(a_0, b_0)$ to this equation for fixed $k$ takes the form $(n, 2n)$ for some positive integer $n$ .

Plugging back into the recursive sequence of solutions, we notice that the solutions alternate between $(n, 2n)$ and $(n, 8n^4-n)$ . Thus, these yield the entire curve of solutions.

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signifance

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signifance

#48 h Sep 19, 2023, 10:30 PM

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Nothing new of a concept here; imo these problems should be less frequent because they're similar to a computational problem where you literally know the approach, you just need to follow through, whereas in problems like sequences or a combo game you need to be creative.

Note that $b=2a$ works, and $(a,b)=(2m,1)$ both work. We claim that the only other solution is $(a,b)=(8m^2-4,2m)$ ; henceforth assume the first two are not true. Then $\frac{a^2}{(2a-b)b^2+1}:=k,b^3k-k < b^3k <\frac{a^2b}{2a-b}<a^2\iff a>b;a^2-2ab^2k+b^3k-k=0\implies(a,b)\Leftrightarrow\left(\frac{b^3k-k}a,b\right)$ yields a smaller solution as long as $a>b$ . If $a\le b,b\ne 2a,a^2\ge2ab^2-b^3=1>0,\text{ then }a^2>b^2(2a-b)\ge0\stackrel{\implies2a-b\ge0}{a\not>b\implies}2a-b=0,$ which we've already accounted for. Plugging back in, we find that the solution set alternates between $(m,2m)\leftrightarrow(m,8m^4-m)$ , which is all the solutions by our Vieta jumping, as desired.

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IAmTheHazard

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IAmTheHazard

#49 h Dec 22, 2023, 1:59 PM

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What is a quadratic

The answer is $(n,2n)$ , $(8n^4-n,2n)$ , $(2n,1)$ only for any $n \in \mathbb{Z}^+$ , which clearly work. Obviously if $b=1$ then $a$ must be even, so suppose $b>1$ .

Rewrite the denominator as $b^2(2a-b)+1$ . For the quantity to be positive, we need $2a-b \geq 0$ . If $2a=b$ then we get the first curve of solutions. Otherwise, I claim that $a>b^2$ . Indeed, if $b \geq a$ then the denominator is at least $b^2+1$ which is too big, and if $\sqrt{a} \leq b \leq a$ then the denominator is at least $(\sqrt{a})^2a+1$ which is also too big.

Now, if $2ab^2-b^3+1 \mid a^2$ then $2ab^2-b^3+1 \mid a^2(2b^2)^2$ , and by the Euclidean algorithm it follows that $2ab^2-b^3+1 \mid (b^3-1)^2$ . Obviously, $2ab^2-b^3+1$ is $1 \pmod{b^2}$ , and since $(b^3-1)^2 \equiv 1 \pmod{b^2}$ as well it follows that $\tfrac{(b^3-1)^2}{2ab^2-b^3+1}$ is a $1 \pmod{b^2}$ positive integer as well. On the other hand, since $a>b^2$ we have
$\frac{(b^3-1)^2}{2ab^2-b^3+1}<\frac{(b^3-1)^2}{2b^4-b^3+1}\leq b^2 \impliedby b^6-2b^3+1\leq 2b^6-b^5+b^2 \iff b^6+2b^3+b^2 \geq b^5+1,$ which is obvious since $b$ is positive integer. Hence we need
$\tfrac{(b^3-1)^2}{2ab^2-b^3+1}=1 \implies b^3(b^3-1)=2ab^2 \implies a=\frac{1}{2}b(b^3-1).$ Plugging this back into the original, for this curve of solutions we would need
$\frac{\frac{1}{4}b^2(b^3-1)^2}{b^3(b^3-1)-b^3+1}=\frac{b^2(b^3-1)^2}{4(b^3-1)^2}=\frac{b^2}{4} \in \mathbb{Z},$ which clearly implies $b=2n$ for some $n \in \mathbb{Z}$ and $a=8n^4-n$ . $\blacksquare$

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shendrew7

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shendrew7

#50 h Apr 28, 2024, 3:35 AM

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Set the expression equal to $k$ . Our equation can be rewritten as
$a^2-(2b^2k)a+(b^3-1)k=0.$
The discriminant of this quadratic is
$\ell^2 := 4b^4k^2-4b^3k+4k = (2b^2k-b)^2 - (b^2-4k).$
This value is at almost always strictly between $(2b^k-b-1)^2$ and $(2b^k-b+1)^2$ , and would produce no solutions unless:

$b=1$ : Substituting into the original expression, we have the solution $\boxed{(2c,1)}$ .
$\ell=\tfrac{b^2}{4}$ : Then $n = \pm (2b^2k-b) = \pm \left(\tfrac{b^4}{4} - \tfrac b2\right)$ , so quadratic formula gives the solutions $\boxed{(c,2c), (8c^2-1,2c)}$ . $\blacksquare$

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Mr.Sharkman

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Mr.Sharkman

#51 h May 14, 2024, 8:31 PM

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The solutions are $(a, b) = (k, 2k), (k, 1), (8k^{4}-k, 2k)$ for positive integers $k.$ Obviously these work. We will show that these are all of the solutions. Take a pair $(a, b)$ which is not of one of the first two forms.

Claim: Such pairs must satisfy $a > b^{2}.$

Assume for the sake of contradiction that $a \le b^{2}.$ We have that
$\frac{a^{2}}{2ab^{2}-b^{3}+1} < \frac{1}{b^{2}} \left(\frac{a^{2}}{2a-b} \right) = \frac{1}{b^{2}} \left(\frac{a}{2}+\frac{b}{4}+\frac{b^{2}}{4(2a-b)} \right).$ Since $2a-b \neq 0,$ we have that
$\frac{b^{2}}{4(2a-b)} \le \frac{b^{2}}{4},$ and $b < b^{2},$ since $b > 1.$ So, we have
$\frac{1}{b^{2}} \left(\frac{a}{2}+\frac{b}{4}+\frac{b^{2}}{4(2a-b)} \right) < \frac{1}{b^{2}} \left( \frac{b^{2}}{2}+\frac{b^{2}}{4}+\frac{b^{2}}{4} \right) = 1,$ so this number cannot be a positive integer, as it is smaller than $1.$

Notice that if
$\frac{a^{2}}{2ab^{2}-b^{3}+1} = k,$ then
$a^{2}-2akb^{2}+(b^{3}-1)k = 0.$ So, now, if $a'$ is the other solution to this equation, $a'$ is an integer (since $a+a^{\prime} = 2kb^{2}$ by Vieta's), and is positive, since $aa^{\prime} = (b^{3}-1)k ,$ again by Vieta's.

Claim: $2a' = b.$

Assume for the sake of contradiction that this is not true, i.e. $(a^{\prime}, b)$ is not of one of the first two forms. (Notice that $b \neq 1,$ so it cannot be the second form.) We get
$a^{\prime} = \frac{(b^{3}-1)k}{a} < \frac{b^{3}k}{a} = \frac{b^{3}a}{2ab^{2}-b^{3}+1} < \frac{ab}{2a-b} < \frac{b}{2} + \frac{\frac{b^{2}}{2}}{2b^{2}-b} \le \frac{b}{2} + \frac{1}{3} < b.$ But, $(a^{\prime}, b)$ is a valid solution set, so $b^{2} < a^{\prime},$ a contradiction.

Now, if we take $b = 2 a^{\prime}$ and $a = \frac{(b^{3}-1)k}{a^{\prime}} = \frac{(8(a^{\prime})^{3}-1)k}{a^{\prime}},$ and replace $\ell = a^{\prime},$ we get that
$a = \frac{(8\ell^{3}-1) \ell^{2}}{\ell} = 8\ell^{4}-\ell,$ giving us our third solution.

This post has been edited 1 time. Last edited by Mr.Sharkman, May 14, 2024, 8:32 PM

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Markas

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Markas

#52 h Jul 13, 2024, 2:47 PM

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Let $\frac{a^2}{2ab^2 - b^3 + 1} = k$ , where k is an integer. We get the quadratic $a^2 - (2b^2k)a + b^3k - k = 0$ . We have that $D = 4b^4k^2 + 4(k - b^3k) = (2b^2k - b)^2 - b^2 + 4k$ . We want a and b to be integers $\Rightarrow$ $D = n^2 = (2b^2k - b)^2 - b^2 + 4k$ .

Case 1: $b^2 - 4k = 0$

From the quadratic $a_1 = \frac{2b^2k + (2b^2k - b)}{2}$ and $a_2 = \frac{2b^2k - (2b^2k - b)}{2}$ $\Rightarrow$ $a_1 = \frac{4b^2k - b}{2} = \frac{b^4 - b}{2}$ for even b. Let b = 2c and from that we get the solutions $(a,b) = (8x^4-x,2x)$ . Also $a_2 = \frac{b}{2}$ and from that we get the solutions $(a,b) = (x,2x)$ .

Case 2: $b^2 - 4k > 0$

We have that $b^2-4k = (2kb^2-b)^2-n^2 \ge (2kb^2-b)^2-(2kb^2-b-1)^2 = 4kb^2-2b-1 > b^2-4k$ , which obviously is a contradiction $\Rightarrow$ we get no solutions from this case.

Case 3: $b^2 - 4k < 0$

We have $4k-b^2 = n^2-(2kb^2-b)^2 \ge (2kb^2-b+1)^2-(2kb^2-b)^2 = 4kb^2-2b+1$ . If $b \ge 2$ , then $4kb^2-2b+1 > 4k-b^2$ , which leads to contradiction. If b = 1, then $4kb^2-2b+1 \le 4k-b^2$ , and we get $n = 2kb^2-b+1$ . Plugging in $b = 1$ gives $a = 2x$ , from which we get the solutions (a,b) = (2x,1) $\Rightarrow$ all solutions we get are: $(a,b) = (8x^4-x,2x); (x,2x); (2x,1)$ .

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OronSH

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OronSH

#53 h Aug 9, 2024, 4:07 PM • 1 Y

Y by megarnie

We claim solutions are $(2n,1),(n,2n),(8n^4-n,2n).$

First note $2ab^2-b^3+1\mid a^2\implies 2ab^2-b^3+1\mid a(b^3-1),$ and letting $k(2ab^2-b^3+1)=a(b^3-1)$ rearranges to $2akb^2=(a+k)(b^3-1).$ If $b=1$ we get the first solution set $(2n,1)$ so assume $b>1.$ Then $b^2\mid a+k.$ Now we can rearrange our equation to $\frac1a+\frac1k=\frac{2b^2}{b^3-1}.$ Since this is symmetric in $a,k$ we WLOG $a\le k.$ Then $\frac2a\ge\frac{2b^2}{b^3-1}$ so $a\le\frac{b^3-1}{b^2}\implies a\le b-1.$ In particular $a<\frac{b^2}{b+1},$ which rearranges to $b^2-a>ab.$ We also have $k\ge b^2-a,$ so $k>ab.$

This gives us the following estimates: $\frac1a\left(1+\frac1b\right)>\frac1a+\frac1k=\frac{2b^2}{b^3-1}>\frac1a.$ The lower bound rearranges to $a>\frac{b^3-1}{2b^2}>\frac{b-1}2$ and the upper bound becomes $a<\frac{(b^3-1)(b+1)}{2b^3}<\frac{b+1}2.$

Thus $a=\frac b2$ which gives $k=\frac{b^4-b}2.$ Since we assumed $a\le k$ it follows that $a$ can take both of these, and setting $b=2n$ gives our other solution sets $(n,2n),(8n^4-n,2n).$ Checking, these work so we are done.

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Marcos_Vinicius

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Marcos_Vinicius

#54 h Oct 4, 2024, 6:33 PM

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Let $\frac{a^2}{2ab^2 - b^3 + 1} = k \implies$ $a^2 - 2ab^2k + b^3k - k = 0$ Obvious $a \neq b$ , so take $a_1$ the other root and $\begin{cases} a + a_1 = 2b^2k \\ a \cdot a_1 = b^3k - k \end{cases}$
If $a_1 = 0 \implies b = 1$ (cause $k \neq 0$ ) $\implies a = 2k \implies$ $(a, b) = (2k, 1)$
If $a_1 > 0$ , let $(a, b)$ the pair with the smallest sum, so $a_1 \geq a \implies a_1 \geq b^2k$ $\implies (b^3 - 1)k \geq b^2ka \implies$ $b > a$ and $k > 0 \implies$ $2ab^2 - b^3 = (2a - b)b^2 \geq 0$ $\implies 2a \geq b$ . Supose $2a > b \implies \frac{a^2}{2ab^2 - b^3 + 1} \leq$ $\frac{a^2}{b^2 + 1} <$ $\frac{a^2}{a^2 + 1} < 1$ $\implies k < 1$ ABS! Then $2a = b \implies k = a^2$ $\implies (a, b) = (a, 2a) \to (8a^4 - a, 2a)$ (by vieta)
$\boxed{\therefore (a, b) = (2t, 1), (t, 2t), (8t^4 - t, 2t)}$

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asdf334

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asdf334

#55 h Yesterday at 9:34 PM • 1 Y

Y by OronSH

okay vieta jumping

Write $k=\frac{a^2}{2ab^2-b^3+1}$ , which is a positive integer. We can rearrange into
$a^2-(2b^2k)a+k(b^3-1)=0$ and now if $(a,b)$ is a solution to the original problem then so is $\left(\frac{k(b^3-1)}{a},b\right)$ .

At this point let's handle the case where $b=1$ , so we can assume that $\frac{k(b^3-1)}{a}$ is an integer in the future. When $b=1$ , we find $k=\frac{a}{2}$ , so any pair $\boxed{(2n,1)}$ is valid.

Now suppose that we have such a pair of solutions $(a_1,b)$ and $(a_2,b)$ where $a_1$ and $a_2$ are positive. Furthermore, assume that $a_1,a_2\ge b$ . Then we actually find
$a_1^2=k(2a_1b^2-b^3+1)\ge k(b^3+1)$ and likewise for $a_2^2$ . Hence $a_1a_2\ge k(b^3+1)$ , but we also know that $a_1a_2=k(b^3-1)$ . That is a contradiction.

That contradiction came from assuming that $a_1,a_2\ge b$ . Hence we can assume WLOG that $a_1<b$ . Just to keep things simple, let's return back to the problem's original notation, assuming that $a<b$ .

Notice $2ab^2-b^3+1=b^2(2a-b)+1$ . If $2a-b\ne 0$ , then this quantity has magnitude at least $b^2-1$ . Since it cannot be $b^2$ , as then it would not be a factor of $a^2$ , it must then be equal to $b^2-1$ . In that case we still require $\frac{a^2}{b^2-1}=1$ , which is not possible.

Hence $2a=b$ , and one can check that $\boxed{(n,2n)}$ is a solution. Finally, this solution pair corresponds to another pair, which was mentioned earlier as a result of Vieta Jumping. Both solutions have $k=n^2$ , and the other solution pair is
$\left(\frac{k(b^3-1)}{a},b\right)=\boxed{(n(8n^3-1),2n)}$ and that concludes the problem. Nice !! slow progress bit by bit. hope i don't fail usamo i guess guh idk $\blacksquare$

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