Topsy-turvy

Let's first look at $(2)$ , there is no such non-negative integer solutions $(x, y)$ that satisfy $(2)$ .

Let me tell you why...

Assume they do exist, then if we divide both sides by an integer, the remainder on both sides must be the same.

Now let's try dividing both sides by 5, the remainder of an integer when divided by 5 can only be $0$ , $1$ , $2$ , $3$ or $4$ . $j\equiv 0,\,1,\,2,\,3,\,4 \pmod5$ $\begin{aligned} \therefore j^2&\equiv 0,\,1,\,4,\,9,\,16 \\ &\equiv 0,\,1,\,4\pmod 5\end{aligned}$ $\begin{aligned} \therefore j^4&\equiv 0,\,1,\,16 \\ &\equiv 0,\,1\pmod5 \end{aligned}$ Thus, $\begin{aligned} \text {L.H.S.}&=x^4+5y\\&\equiv 0,\,1 \pmod 5\end{aligned}$ $\begin{aligned} \text {R.H.S}&=n^4-2\\&\equiv -2,\,-1\\&\equiv3,\,4 \pmod 5\end{aligned}$ We see that the remainder of the L.H.S. and R.H.S will never be the same, which is a contradiction. So, there are no non-negative integer solutions that satisfy $(2)$ .

Therefore $q=0$ , $p=9$ .

Now let's observe $(1)$ , simplify it to become $5x+4y=n^2\qquad(3)$ We see that this is a linear Diophantine equation with 2 variables. Divide both sides by the smallest coefficient in L.H.S. which is 4. Then move all fractions to the right and all integers to the left, we get $x+y=\frac{n^2-x}{4}$ Since $x+y$ is an integer, $\frac{n^2-x}{4}$ must also be an integer. Let $\frac {n^2-x}{4}=k$ , then, $x=n^2-4k$ Substitute $x=n^2-4k$ into $(3)$ , then we would get $y=5k-n^2$ Since $x$ and $y$ are non-negative, $x\geqslant0$ , $y\geqslant 0$ , we have $4k\leqslant n^2\leqslant 5k$ $\therefore\frac {n^2}{5}\leqslant k \leqslant \frac {n^2}{4}\qquad(4)$

The number of integer solutions $k$ that satisfy $(4)$ is $\left\lfloor\frac{n^2}{4}\right\rfloor-\left\lceil\frac {n^2}{5}\right \rceil+1$ . Since the number of non-negative integer solutions that satisfy $(3)$ is equal to the number of integer solutions $k$ that satisfy $(4)$ , $p=\left\lfloor\frac{n^2}{4}\right\rfloor-\left\lceil\frac {n^2}{5}\right \rceil+1=9$ $\therefore\left \lfloor \frac {n^2}{4} \right \rfloor-\left \lceil \frac {n^2}{5}\right \rceil=8$ Because $n$ is a positive integer, we can solve this via trial and error by plugging in various values of $n$ , and it is not hard to show that $n=13$ is the only solution.

Thus, the value of $n$ is $13$ .