# Topsy-turvy

**Algebra**Level 4

Let's begin with an equation: \[x_1+y_1=n\] (\(n\) is a positive integer)

That's too boring, so let's add some numbers, let's make 5, 4, 4 and 2 as the main protagonist: \[5x_2+4y_2=n^{4-2}~~~~~~(1)\] Unfortunately, someone came and messed up the numbers, making a new equation: \[x_3^4+5y_3=n^4-2~~~~~~(2)\] If there are \(p\) ordered pairs of non-negative integer solutions \((x,y)\) that satisfies \((1)\), there are \(q\) ordered pairs of non-negative integer solutions \((x,y)\) that satisfies \((2)\), and \(\vert p-q\vert=9\), what is the value of \(n\)?

This is one part of 1+1 is not = to 3.