Thank you to Maria from Hills Road Sixth Form College, Eleanor from The King's School in Macclesfield, Abi, and Rhombusta for submitting solutions to this problem!

Here's Abi's solution:

For equations of the form $(ax^2+bx+c)^{(dx^2+ex+f)}=1$, where $a,b,c,d,e,f$ and $x$ are real, there are three possibilities:

$i)$

Case $1$: $x^2-7x+11=1$ factorises as $(x-5)(x-2)=0$ so $x=5$ or $x=2$.

Case $2$: $x^2-11x+30=0$ factorises as $(x-5)(x-6)=0$, so $x=6$ or $x=5$.

Case $3$: $x^2-7x+11=-1$ factorises as $(x-3)(x-4)=0$. As $3^2-11(3)+30=6$ and $4^2-11(4)+30=2$ are both even, $x=3$ or $x=4$ are solutions.

So the solutions are $x=2,3,4,5$ and $6$.

$ii)$

Case $1$: $2-x^2=1$ factorises as $(1-x)(1+x)=0$ so $x=1$ or $x=-1$.

Case $2$: $x^2-3 \sqrt{2}x+4=0$ factorises as $(x-2 \sqrt{2})(x- \sqrt{2})$, so $x=2 \sqrt{2}$ or $x = \sqrt{2}$.

Case $3$: $2-x^2=-1$ factorises as $(\sqrt{3}-x)(\sqrt{3}-x)$, but as $\sqrt{3}^2-3 \sqrt{2} \sqrt{3} +4$ and $ \sqrt{3}^2-3 \sqrt{2} \sqrt{3} + 4$ are not even integers, these are invalid.

So the solutions are $x=-1,1, \sqrt{2}$ and $2 \sqrt{2}$.

Here's Abi's solution:

For equations of the form $(ax^2+bx+c)^{(dx^2+ex+f)}=1$, where $a,b,c,d,e,f$ and $x$ are real, there are three possibilities:

- $ax^2+bx+c=1$, as $1^n=1$ for any $n$
- $dx^2+ex+f=0$, as $n^0=1$ for any $n$
- $ax^2+bx+c=-1$, and $dx^2+ex+f$ is an even integer, as $(-1)^{(2n)}=1$ for any integer $n$.

$i)$

Case $1$: $x^2-7x+11=1$ factorises as $(x-5)(x-2)=0$ so $x=5$ or $x=2$.

Case $2$: $x^2-11x+30=0$ factorises as $(x-5)(x-6)=0$, so $x=6$ or $x=5$.

Case $3$: $x^2-7x+11=-1$ factorises as $(x-3)(x-4)=0$. As $3^2-11(3)+30=6$ and $4^2-11(4)+30=2$ are both even, $x=3$ or $x=4$ are solutions.

So the solutions are $x=2,3,4,5$ and $6$.

$ii)$

Case $1$: $2-x^2=1$ factorises as $(1-x)(1+x)=0$ so $x=1$ or $x=-1$.

Case $2$: $x^2-3 \sqrt{2}x+4=0$ factorises as $(x-2 \sqrt{2})(x- \sqrt{2})$, so $x=2 \sqrt{2}$ or $x = \sqrt{2}$.

Case $3$: $2-x^2=-1$ factorises as $(\sqrt{3}-x)(\sqrt{3}-x)$, but as $\sqrt{3}^2-3 \sqrt{2} \sqrt{3} +4$ and $ \sqrt{3}^2-3 \sqrt{2} \sqrt{3} + 4$ are not even integers, these are invalid.

So the solutions are $x=-1,1, \sqrt{2}$ and $2 \sqrt{2}$.