unusual integer equation

**abhishekkgp** · July 2nd 2011, 06:40 PM

Find all the positive integers $x, \, y$ such that $x^y=(y+1)^4$ .

I could show that no solution is there if $5 \leq y+1<x$ .
But the remaining cases still bug me.
any help.

**TKHunny** · July 2nd 2011, 06:57 PM

Seems like there is quite a collection for y = 0.

**abhishekkgp** · July 2nd 2011, 09:54 PM

Originally Posted by TKHunny

Seems like there is quite a collection for y = 0.

there is. but we are concerned with only positive integers.

**Also sprach Zarathustra** · July 2nd 2011, 11:26 PM

Originally Posted by abhishekkgp

Find all the positive integers $x, \, y$ such that $x^y=(y+1)^4$ .

I could show that no solution is there if $5 \leq y+1<x$ .
But the remaining cases still bug me.
any help.

y=4
x=y+1

(5,4) is a solution. I think it is the only one.(for x,y>0)

**TKHunny** · July 3rd 2011, 02:31 AM

Originally Posted by abhishekkgp

there is. but we are concerned with only positive integers.

Always with the missing ONE word!

**abhishekkgp** · July 7th 2011, 02:49 AM

Originally Posted by Also sprach Zarathustra

y=4
x=y+1

(5,4) is a solution. I think it is the only one.(for x,y>0)

just found another one (x,y)=(3,8)

**melese** · July 7th 2011, 08:09 AM

Originally Posted by abhishekkgp

Find all the positive integers $x, \, y$ such that $x^y=(y+1)^4$ .

I could show that no solution is there if $5 \leq y+1<x$ .
But the remaining cases still bug me.
any help.

Write (a) $x^{y/4}=y+1$ . Clearly, $x\geq2$ and so $y+1=x^{y/4}\geq2^{y/4}$ . Note that since $x^y$ is a fourth power, then $y/4$ must be an integer.*

It can be shown# that $2^n>4n+1$ for any integer $n\geq5$ , so $4n+1\leq2^n$ implies $n<5$ . Now, taking $n=y/4$ we have $y/4<5$ .

The only possible value for $y$ are $4, 8, 12,$ and $,16$ . The corresponding equations for (a) are $x=5$ , $x^2=9$ , $x^3=13$ , and $x^4=17$ . Only the first two are solvable. Therefore, the only solutions to the originial equation are those that Also sprach Zarathustra and abhishekkgp gave.

*Not necessary for the solution, only to reduce the number cases.
#If $n=5$ , then $2^5>4\cdot5+1$ . Adding $4$ gives $4+2^n<2^n+2^n=2^{n+1}$ and $4n+1+4=4(n+1)+1$ . Using $2^n>4n+1$ we have $2^{n+1}>4(n+1)+1$ .

**abhishekkgp** · July 7th 2011, 08:18 AM

Originally Posted by melese

Write (a) $x^{y/4}=y+1$ . Clearly, $x\geq2$ and so $y+1=x^{y/4}\geq2^{y/4}$ . Note that since $x^y$ is a fourth power, then $y/4$ must be an integer.*

i don't think this is correct for if we take $x=4, y=2$ then $x^y=2^4$ is still a fourth power but $y/4$ is not an integer.

**melese** · July 7th 2011, 11:12 AM

Originally Posted by abhishekkgp

i don't think this is correct for if we take $x=4, y=2$ then $x^y=2^4$ is still a fourth power but $y/4$ is not an integer.

I cannot be sure I'm not deceived again.

Since $y>0$ , $x\geq2$ . So, $2^y\leq x^y$ and by the Binomial Theorem, $(1+y)^4<1+y^4$ . Therefore, $2^y\leq y^4$ .

But $2^y>y^4$ for $y>16$ (by induction or otherwise). It follows that $y\leq16$ .

The possible equations are:
$x^1=(1+1)^4; x^2=(2+1)^4; x^3=(3+1)^4; x^4=(4+1)^4;...; x^{15}=(15+1)^4; x^{16}=(y+1)^4$ . Then the only solutions I've found are $(16,1), (9,2), (5,4), (3,8)$ .

**Goku** · July 7th 2011, 12:22 PM

could you try y=4z and x=(4z+1)^1/z, i don,t think that they are any more +ve integer solutions, I wrote a program and as y goes to infinity x approaches 1, any ideas..

**abhishekkgp** · July 7th 2011, 07:05 PM

Originally Posted by melese

I cannot be sure I'm not deceived again.

Since $y>0$ , $x\geq2$ . So, $2^y\leq x^y$ and by the Binomial Theorem, $(1+y)^4<1+y^4$ . Therefore, $2^y\leq y^4$ .

But $2^y>y^4$ for $y>16$ (by induction or otherwise). It follows that $y\leq16$ .

The possible equations are:
$x^1=(1+1)^4; x^2=(2+1)^4; x^3=(3+1)^4; x^4=(4+1)^4;...; x^{15}=(15+1)^4; x^{16}=(y+1)^4$ . Then the only solutions I've found are $(16,1), (9,2), (5,4), (3,8)$ .

i had known that there are only for solutions to that equations.. i have not yet reviewed your solution but m sure u r right this time!

**abhishekkgp** · July 7th 2011, 07:12 PM

Originally Posted by melese

$(1+y)^4<1+y^4$ . .

but its the other way around. Even if this is a typo i am not getting how you got y<16..

**melese** · July 8th 2011, 12:25 AM

Originally Posted by abhishekkgp

but its the other way around. Even if this is a typo i am not getting how you got y<16..

That was not a typo but a serious mistake. We have $2^y\leq x^y=(1+y)^4$ .

For $y>16$ , the inequality $2^y>(1+y)^4$ can be shown by induction on $y$ . We can check that $2^{17}>(1+17)^4$ . Assuming that $2^y>(1+y)^4$ multiply both sides by $2$ . Then $2^{y+1}>(1+y)^4\cdot2$ . Now it's left to show that $(1+y)^4\cdot2>(2+y)^4$ . Note that $\displaystyle\frac{(2+y)^4}{(1+y)^4}=(\frac{2+y}{1 +y})^4=(1+\frac{1}{1+y})^4$ .
Since $(1+\frac{1}{1+y})^4$ gets smaller as $y>16$ gets larger it follows that $(1+\frac{1}{1+y})^4<(1+\frac{1}{1+16})^4<2$ and therefore $\frac{(2+y)^4}{(1+y)^4}<2$ - this completes the inductive part.

Since we want $2^y\leq(1+y)^4$ , we must take $y\leq16$ . Otherwise $y>16$ and the inequality $2^y>(1+y)^4$ holds.

**abhishekkgp** · July 8th 2011, 01:43 AM

Originally Posted by melese

That was not a typo but a serious mistake. We have $2^y\leq x^y=(1+y)^4$ .

For $y>16$ , the inequality $2^y>(1+y)^4$ can be shown by induction on $y$ . We can check that $2^{17}>(1+17)^4$ . Assuming that $2^y>(1+y)^4$ multiply both sides by $2$ . Then $2^{y+1}>(1+y)^4\cdot2$ . Now it's left to show that $(1+y)^4\cdot2>(2+y)^4$ . Note that $\displaystyle\frac{(2+y)^4}{(1+y)^4}=(\frac{2+y}{1 +y})^4=(1+\frac{1}{1+y})^4$ .
Since $(1+\frac{1}{1+y})^4$ gets smaller as $y>16$ gets larger it follows that $(1+\frac{1}{1+y})^4<(1+\frac{1}{1+16})^4<2$ and therefore $\frac{(2+y)^4}{(1+y)^4}<2$ - this completes the inductive part.

Since we want $2^y\leq(1+y)^4$ , we must take $y\leq16$ . Otherwise $y>16$ and the inequality $2^y>(1+y)^4$ holds.

that's great!!

Thread: unusual integer equation

LinkBack

Thread Tools

Display

unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Re: unusual integer equation

Similar Math Help Forum Discussions

Integer Soultion to an equation...

Integer solutions to equation p^3=q^2

integer solutions of equation

Integer equation

Unusual equation