Show that is the square of an integer

[Joppy]

I couldn't find this problem anywhere else on the forum so I thought I'd post it. If however, I am duplicating, mods feel free to remove the post :p.

No doubt many of you know it already, but I found it quite interesting.

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $\frac{a^2 + b^2}{ab + 1}$ is the square of an integer.

 

[topsquark]

I couldn't find this problem anywhere else on the forum so I thought I'd post it. If however, I am duplicating, mods feel free to remove the post :p.

No doubt many of you know it already, but I found it quite interesting.

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $\frac{a^2 + b^2}{ab + 1}$ is the square or an integer.

This one's infuriating. I love it!

-Dan

 

[Joppy]

This one's infuriating. I love it!

-Dan

So many layers!

 

[Albert1]

I couldn't find this problem anywhere else on the forum so I thought I'd post it. If however, I am duplicating, mods feel free to remove the post :p.

No doubt many of you know it already, but I found it quite interesting.

Let $a$ and $b$ be positive integers such that $ab + 1$ divides $a^2 + b^2$. Show that $\frac{a^2 + b^2}{ab + 1}$ is the square of an integer.

my solution:

Spoiler

if $a=b^3$
then $\dfrac {a^2+b^2}{ab+1}=\dfrac {b^6+b^2}{b^4+1}=b^2$
if $b=a^3$
then $\dfrac {a^2+b^2}{ab+1}=\dfrac {a^2+a^6}{a^4+1}=a^2$
$\therefore ab+1$ divides $a^2+b^2$ we may set $a=b^3,or\,\, b=a^3$
(here $a,b \in N$)

 

[mrtwhs]

This is problem #3 day 2 of the 1988 IMO.

 

[Joppy]

This is problem #3 day 2 of the 1988 IMO.

Thought it was no. 6.

 

[topsquark]

my solution:

Spoiler

if $a=b^3$
then $\dfrac {a^2+b^2}{ab+1}=\dfrac {b^6+b^2}{b^4+1}=b^2$
if $b=a^3$
then $\dfrac {a^2+b^2}{ab+1}=\dfrac {a^2+a^6}{a^4+1}=a^2$
$\therefore ab+1$ divides $a^2+b^2$ we may set $a=b^3,or\,\, b=a^3$
(here $a,b \in N$)

How do we know that this is the most general solution (or that others don't exist)? I can't think of how to prove that one way or another.

-Dan

 

[Joppy]

How do we know that this is the most general solution (or that others don't exist)? I can't think of how to prove that one way or another.

-Dan

I think Albert has stumbled upon one set of solutions, but there are many more in similar forms.. But! I'll have to check myself first when I get a chance.

 

[Albert1]

How do we know that this is the most general solution (or that others don't exist)? I can't think of how to prove that one way or another.

-Dan

Spoiler

since we are given :$a^2+b^2 $ is a multiple of $ab+1$
we must find the relation between $a$ and $b$ as my example $a=b^3$ or $b=a^3$
may be we can find different link between $a,b$ and check the result
the following steps will help us finding other solutions:
if $a>b$
take $\dfrac {a^2+b^2}{ab+1}=2^2=4=k^2----(*)$ for instance
(1) set $a=b^3$,we get $a=8,b=2$
(2) put $b=8 $ to $(*)$ we get $a=2$ ,or $a=30$ another solution $ (a,b)=(30,8) $ is found
(3) put $b=30$ to $(*)$ we get $a=8$ or $a=112$ again $(a,b)=(112,30)$ is found
(4) put $b=112 $ to $(*)$ we get $a=30$ ,or $a=418$ another solution $ (a,b)=(418,112) $ is found
continue these steps until no other solution found then next $k$

 

[Albert1]

Spoiler

since we are given :$a^2+b^2 $ is a multiple of $ab+1$
we must find the relation between $a$ and $b$ as my example $a=b^3$ or $b=a^3$
may be we can find different link between $a,b$ and check the result
the following steps will help us finding other solutions:
if $a>b$
take $\dfrac {a^2+b^2}{ab+1}=2^2=4=k^2----(*)$ for instance
(1) set $a=b^3$,we get $a=8,b=2$
(2) put $b=8 $ to $(*)$ we get $a=2$ ,or $a=30$ another solution $ (a,b)=(30,8) $ is found
(3) put $b=30$ to $(*)$ we get $a=8$ or $a=112$ again $(a,b)=(112,30)$ is found
(4) put $b=112 $ to $(*)$ we get $a=30$ ,or $a=418$ another solution $ (a,b)=(418,112) $ is found
continue these steps until no other solution found then next $k$

The general solutions of :$\dfrac {a^2+b^2}{ab+1}=k^2-----(*)\,\,\,(a,b,k\in N)$

Spoiler

because of symmetry we let $a>b$
from previous solution we have $ b=k,a=b^3=k^3 $
put $b=k^3$ to $(*)$
we find another solution of $a$
$\dfrac {a^2+k^6}{ak^3+1}=k^2$
$\rightarrow a^2-ak^5+k^6-k^2=0$
$a=k,\,\, or \,\ ,a=k^5 - k$

 

[kaliprasad]

The general solutions of :$\dfrac {a^2+b^2}{ab+1}=k^2-----(*)\,\,\,(a,b,k\in N)$

Spoiler

because of symmetry we let $a>b$
from previous solution we have $ b=k,a=b^3=k^3 $
put $b=k^3$ to $(*)$
we find another solution of $a$
$\dfrac {a^2+k^6}{ak^3+1}=k^2$
$\rightarrow a^2-ak^5+k^6-k^2=0$
$a=k,\,\, or \,\ ,a=k^5 - k$

Albert,
you are digressing

Spoiler

you are finding general solution instead of proving that it is perfect square

 

[Albert1]

Albert,
you are digressing

Spoiler

you are finding general solution instead of proving that it is perfect square

may be a little bit digressing,but from this I can deduce that it is perfect square:

Spoiler

$let :\dfrac{a^2+b^2}{ab+1}=y\\
a^2-yab+b^2-y=0\\
a=\dfrac{yb\pm\sqrt{y^2b^2-4b^2+4y}}{2}$
since $a,b\in N$
$y^2b^2-4b^2+4y$ must be perfect square
this can only be done with $b=y\sqrt y=k^3$
$a=\dfrac {y^2\sqrt y \pm (y^2\sqrt y-2\sqrt y)}{2}=y^2\sqrt y-\sqrt y =k^5-k\,\, or \,\, a=\sqrt y=k$
as given from general solution
here $(y=k^2)$ must be perfect square

 

[kaliprasad]

may be a little bit digressing,but from this I can deduce that it is perfect square:

Spoiler

$let :\dfrac{a^2+b^2}{ab+1}=y\\
a^2-yab+b^2-y=0\\
a=\dfrac{yb\pm\sqrt{y^2b^2-4b^2+4y}}{2}$
since $a,b\in N$
$y^2b^2-4b^2+4y$ must be perfect square
this can only be done with $b=y\sqrt y=k^3$
$a=\dfrac {y^2\sqrt y \pm (y^2\sqrt y-2\sqrt y)}{2}=y^2\sqrt y-\sqrt y =k^5-k\,\, or \,\, a=\sqrt y=k$
as given from general solution
here $(y=k^2)$ must be perfect square

Spoiler

This is a mixup between arithmetic and algebra. for example $b^2+5$ is not a perfect square in algebra but is it in arthmetic when $b=2$ as $2^2+5= 3^2$

 

[Albert1]

Spoiler

This is a mixup between arithmetic and algebra. for example $b^2+5$ is not a perfect square in algebra but is it in arthmetic when $b=2$ as $2^2+5= 3^2$

Spoiler

note $a,b\in N$ and $ab+1$ divides $a^2+b^2$
$y^2b^2-4b^2+4y=(yb-2\sqrt y)^2$ is perfect square
this implies $b=y\sqrt y \in N$
$\therefore y$ must be perfect square
so $y=k^2, b=k^3$