1. ## unusual integer equation

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

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

2. ## Re: unusual integer equation

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

3. ## Re: unusual integer equation

Originally Posted by TKHunny
Seems like there is quite a collection for y = 0.
there is. but we are concerned with only positive integers.

4. ## Re: unusual integer equation

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

I could show that no solution is there if $\displaystyle 5 \leq y+1<x$.
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)

5. ## Re: unusual integer equation

Originally Posted by abhishekkgp
there is. but we are concerned with only positive integers.
Always with the missing ONE word!

6. ## Re: unusual integer equation

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)

7. ## Re: unusual integer equation

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

I could show that no solution is there if $\displaystyle 5 \leq y+1<x$.
But the remaining cases still bug me.
any help.
Write (a)$\displaystyle x^{y/4}=y+1$. Clearly, $\displaystyle x\geq2$ and so $\displaystyle y+1=x^{y/4}\geq2^{y/4}$. Note that since $\displaystyle x^y$ is a fourth power, then $\displaystyle y/4$ must be an integer.*

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

The only possible value for $\displaystyle y$ are $\displaystyle 4, 8, 12,$ and $\displaystyle ,16$. The corresponding equations for (a) are$\displaystyle x=5$, $\displaystyle x^2=9$, $\displaystyle x^3=13$, and $\displaystyle x^4=17$. 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 $\displaystyle n=5$, then $\displaystyle 2^5>4\cdot5+1$. Adding $\displaystyle 4$ gives $\displaystyle 4+2^n<2^n+2^n=2^{n+1}$ and $\displaystyle 4n+1+4=4(n+1)+1$. Using $\displaystyle 2^n>4n+1$ we have $\displaystyle 2^{n+1}>4(n+1)+1$.

8. ## Re: unusual integer equation

Originally Posted by melese
Write (a)$\displaystyle x^{y/4}=y+1$. Clearly, $\displaystyle x\geq2$ and so $\displaystyle y+1=x^{y/4}\geq2^{y/4}$. Note that since $\displaystyle x^y$ is a fourth power, then $\displaystyle y/4$ must be an integer.*
i don't think this is correct for if we take $\displaystyle x=4, y=2$ then $\displaystyle x^y=2^4$ is still a fourth power but $\displaystyle y/4$ is not an integer.

9. ## Re: unusual integer equation

Originally Posted by abhishekkgp
i don't think this is correct for if we take $\displaystyle x=4, y=2$ then $\displaystyle x^y=2^4$ is still a fourth power but $\displaystyle y/4$ is not an integer.
I cannot be sure I'm not deceived again.

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

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

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

10. ## Re: unusual integer equation

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..

11. ## Re: unusual integer equation

Originally Posted by melese
I cannot be sure I'm not deceived again.

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

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

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

12. ## Re: unusual integer equation

Originally Posted by melese
$\displaystyle (1+y)^4<1+y^4$. .
but its the other way around. Even if this is a typo i am not getting how you got y<16..

13. ## Re: unusual integer equation

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 $\displaystyle 2^y\leq x^y=(1+y)^4$.

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

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

14. ## Re: unusual integer equation

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

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

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