Strange diophantine equation and one solution

Question

I invented a diophantine equation $$ (x^y)^{x}+y^{x^{y}}=(x+y)^y\times(10(x+y)+y) $$ based on the fact that $$ (2^3)^{2}+3^{2^{3}}=(2+3)^3\times(10\times 5+3) =6625. $$ I am pretty sure that there are no other integer solutions, unless you somehow define $0^0$ to be 0. Is there a way to prove it? How could one tackle something like this, maybe with some sort of monotonicity argument.

Answer 1

Solving this over the integers is a matter of appropriately distinguishing cases and then applying standard arguments. As far as I can tell there is nothing particularly challenging or interesting, but there are a lot of cases to cover.

Let $x$ and $y$ be integers such that $$x^{xy}+y^{(x^y)}=(x+y)^y(10x+11y).\tag{1}$$

The case $xy=0$.

If $y=0$ we simply get $$x^0+0^{x^0}=x^0(10x+11\cdot0)=10x.$$ With the convention that $x^0=1$ for all $x$ we find that $10x=1$, a contradiction. Only with the uncommon convention that $0^0=0$ we get $(x,y)=(0,0)$ as the only solution with $y=0$.

Similarly, if $x=0$ then we simply get $$0^0+y^{(0^y)}=y^y(10\cdot0+11y)=11y^{y+1}.$$ Regardless of the convention for $0^0$ we get no new solution with $x=0$.

The case $|x|=1$.

Plugging $x=1$ into $(1)$ yields $$1+y=(y+1)^y(11y+10).$$ Clearly $y=-1$ yields the solution $(x,y)=(1,-1)$. If $y\neq-1$ then dividing by $y+1$ yields $$1=(y+1)^{y-1}(11y+10)=(y+1)^{y-1}(11(y+1)-1).$$ There is clearly no solution with $y>1$, and for $y<0$ we get $$(y+1)^{1-y}=11(y+1)-1,$$ where the left hand side is an integer. This is clearly impossible.

Plugging $x=-1$ into $(1)$ yields $$(-1)^{-y}+y^{(-1)^y}=(y-1)^y(11y-10),$$ If $y$ is even then the left hand side simplifies to $1+y$, which is an odd integer. If $y>0$ then the right hand side is an even integer, a contradiction. If $y<0$ then $y-1$ divides $$11y-10=11(y-1)+1,$$ and hence $y-1$ divides $1$, again a contradiction.

If $y$ is odd then multiplying through by $y$ yields $$-y+1=y(y-1)^y(11y-10).$$ If $y<0$ then for the right hand side to be an integer $y-1$ must divide $11y-10$, which we already saw was impossible. Then $y>0$ and then we see that $y$ divides $1$, yielding the solution $(x,y)=(-1,1)$.

The case $x+y=0$.

If $x=-y$ then the right hand side of $(1)$ vanishes, and so $$x^{(-x^2)}+(-x)^{(x^{-x})}=0.$$ Multiplying by $x^{(x^2)}$ then yields $$1+(-1)^{(x^{-x})}x^{x^2+x^{-x}}=0.$$ From the previous cases we can assume that $|x|>1$, and so it follows that $x^2+x^{-x}=0$, or equivalently that $x^{x+2}=-1$. But then $x=-1$, a contradiction.

The case $10x+11y=0$.

If $10x+11y=0$ then the right hand side of $(1)$ vanishes, and $x=-11z$ and $y=10z$ for some integer $z$, so $$(-11z)^{-110z^2}+(10z)^{\left((-11z)^{10z}\right)}=0.$$ The second term is an integer because the base $10z$ is an integer, and the exponent $(-11z)^{10z}$ is a positive integer because $10z$ is even. The first term is not an integer because the base $-11z$ is an integer, and the exponent $-110z^2$ is negative. This is a contradiction.

The case $x,y<0$.

If $x,y<0$, then multiplying by $(x+y)^{|y|}$ and rearranging a bit we find that $$(x+y)^{|y|}y^{(x^y)}=(10x+11y)-(x+y)^{|y|}x^{xy},$$ where all expressions on the right hand side are integers, because $xy>0$. Then the left hand side is also an integer, and of course $(x+y)^{|y|}$ is also an integer, so $y^{(x^y)}$ is rational. Here we are taking the $x^{|y|}$-th root of $y$. For this to be a rational $y$ must be a perfect $x^{|y|}$-th power, and moreover $x$ must be odd because $y<0$.

Of course if $x<-1$ then $|y|<\left|x^{|y|}\right|$ and so $y$ cannot be a perfect $x^{|y|}$-th power. We see that $x=-1$ and plugging this back into $(1)$ and multiplying by $y$ yields $$(-1)^yy+y^{1+(-1)^y}=y(y-1)^y(11y-10).$$ Here the left hand side is an integer, hence so is the right hand side. Of course $y-1$ is coprime to $y$ so this shows that $(y-1)^y$ divides $11y-10$. But $$11y-10=11(y-1)+1,$$ so it follows that $y-1$ divides $1$, which is impossible as $y<0$. This shows that there are no integral solutions with $x,y<0$.

The case $x<0$ and $y>0$.

If $x<0$ and $y>0$ then the right hand side of $(1)$ is an integer. Considering the earlier case $x=-1$ we may assume that $x<-1$, so that $x^{xy}$ is not an integer. Then $y^{(x^y)}$ is not an integer, which means that $y$ is odd and greater than $1$. Clearing denominators by multiplying by the integer $x^{|x|y}y^{(|x|^y)}$ we find that $$y^{(|x|^y)}+x^{|x|y}=x^{|x|y}y^{(|x|^y)}(x+y)^y(10x+11y).$$ We can rearrange this to get $$\Big((x+y)^y(10x+11y)x^{|x|y}-1\Big)\Big((x+y)^y(10x+11y)y^{(|x|^y)}-1\Big)=1,$$ where the two factors on the left hand side are integers, and hence they are equal. This means that either $x+y=0$ or $10x+11y=0$ or $x^{|x|y}=y^{(|x|^y)}$. We have already covered the first two cases, so it remains to solve $$x^{|x|y}=y^{(|x|^y)}.$$ The right hand side positive because $y$ is positive. Because $x$ is negative and $y$ is odd, it follows that $x$ is even. Then $x^{|x|y}$ is also even, and so the left hand side is even. But we already saw that $y$ is odd, a contradiction.

The case $x>0$ and $y<0$.

If $x>0$ and $y<0$ then the right hand side of $(1)$ is a rational number, and so is $x^{xy}$. Then also $y^{(x^y)}$ is rational, which implies that $y$ is a perfect $x^{|y|}$-th power. It follows that $|y|=1$ because $$|y|<2^{(2^|y|)}<2^{(x^{|y|})},$$ because we may assume that $x>1$, as we have already dealt with the case $|x|=1$ before. Then $y=-1$ and plugging this into $(1)$ yields $$x^{-1}+(-1)^{(x^{-1})}=(x-1)^{-1}(10x-11).$$ Multiplying by $x$ then yields $$1+(-1)^{(x^{-1})}x=x(x-1)^{-1}(10x-11),$$ where the left hand side is an integer. Then $x-1$ divides $x(10x-11)$, and so $x=2$. But then $(-1)^{(x^{-1})}$ is not a rational number, whereas all other terms are rational, a contradiction.

The case $x,y\geq3$.

If $x,y\geq3$ then $x^x\geq3x^2$ and $x^y\geq3y^2$. Let $z:=\max(x,y)$ so that $x+y\leq 2z$ and $10x+11y\leq 11z$. Then \begin{eqnarray*} (11z)(2z)^y&\geq&(x+y)^y(10x+11y)\\ &=&(x^x)^y+y^{(x^y)}\\ &\geq&(3x^2)^y+y^{(3y^2)}\\ &=&(3x^2)^y +(y^{3y})^y\\ &\geq&(3x^2)^y +(3y^2)^y\\ &\geq&(3z^2)^y. \end{eqnarray*} Dividing by $(2z)^y$ and using the fact that $y,z\geq3$ we find that $$11z\geq\left(\frac32z\right)^y=z\left(\frac32\right)^yz^{y-1}\geq\frac{27}{8}\cdot9,$$ a contradiction. So either $x<3$ or $y<3$.

The remaining cases $x=2$ or $y\in\{1,2,3\}$.

Plugging in $y=1$ yields $$x^x+1=(x+1)(10x+11),$$ and reducing mod $x$ shows that $1\equiv11\pmod{x}$ and so $x$ divides $10$. A quick check yields no integral solutions.

Plugging in $y=2$ yields $$x^{2x}+2^{(x^2)}=(x+2)^2(10x+22).$$ Of course the left hand side grows much faster than the right hand side; for all $x\geq2$ the left hand side more than doubles as $x$ increases by $1$, whereas the right hand side is a cubic in $x$. A quick check for $x=2,3,4$ yields no integral solutions, and for $x=4$ the left hand side is already greater than the right hand side.

Plugging in $y=3$ yields $$x^{3x}+3^{(x^3)}=(x+3)^3(10x+33).$$ Again the left hand side grows much faster than the right hand side for all $x\geq2$. You already found that equality holds for $x=2$, yielding the solution $(x,y)=(2,3)$, and hence there are no other solutions with $y=3$.

Plugging in $x=2$ yields $$2^{2y}+y^{(2^y)}=(y+2)^y(11y+20).$$ Again the left hand side grows much faster than the right hand side for all $y\geq2$. We have already found equality for $y=3$, and so there are no other solutions with $x=2$.

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This shows that $(x,y)=(2,3)$ is the unique solution in positive integers, and that the only other integral solutions are $(x,y)=(1,-1)$ and $(x,y)=(1,-1)$, and perhaps $(x,y)=(0,0)$ with the convention that $0^0=0$.