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 $\displaystyle 2^{n+1}>4(n+1)+1$.
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 $\displaystyle (16,1), (9,2), (5,4), (3,8)$.
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.