given N and M count the number of solutions to some diophantine equations, prove that $N>M.$ 
Let $N$ be the number of integer solutions to the equation $x^3 - y^3 = z^4-w^4$ with the property that $0\leq x,y,z,w\leq 2022^{2022}$. Let $M$ be the number of integer solutions to the equation $x^3-y^3 = z^4-w^4+1$ with $0\leq x,y,z,w\leq 2022^{2022}$. Prove that $N>M.$


This is problem 8 from this problem set

Let $a_i$ denote the number of solutions to the equation $s^3 + t^4 = i$ where $0\leq s,t\leq 2022^{2022}$. The maximum value of $i$ for which $a_i > 0$ is $A := 2022^{6066} + 2022^{8088}.$ Let $f(x,y) = x^3 + y^4$ for real numbers x and y. $N$ counts the number of integer solutions to the equation $x^3 + w^4 = y^3 + z^4$ where $0\leq x,y,z,w\leq 2022^{2022}.$ For each $0\leq i\leq A,$ there are $a_i$ pairs $(x,y)$ with $f(x,y)=i$. For each pair $(x,w)$, there are $a_{x^3 + w^4}$ pairs $(y,z)$ so that $f(y,z) = f(x,w)$. Hence $N$ must equal $\sum_{i=0}^A a_i^2$ (so $a_i^2$ is the number of solutions $x,y,z,w$ to the equation with $f(x,w)=f(y,z)=i$). On the other hand, for a pair $(y,z)$ with $f(y,z) = i$, there are $a_{i+1} $ pairs $(x,w)$ with $f(x,w)=i+1$. Thus, $M$ equals $\sum_{i=0}^A a_i a_{i+1}$. Hence to show that $N > M,$ it suffices to show that $\sum_{i=0}^A a_i(a_i-a_{i+1}) > 0$. But I'm not sure how to show the latter inequality. It is clearly not always true that $a_i \ge a_{i+1},$ as this does not hold for $i = 0.$ $a_2 = 1, a_i = 0$ for $3\leq i\leq 7, a_8 = 1, a_9 = 1, a_i = 0$ for $10\leq i\leq 15, a_{16} = 1, a_{17} = 1, a_i = 0$ for $18\leq i\leq 23, a_{24} = 1, a_{25} = 0, a_{26}=0,a_{27}=1, a_{28} = 1.$ There's no noticeable pattern.

Continuing from what you've provided, first note that $a_0 = 1$ and, as you stated about how $A$ is defined, we have $a_{A+1}=0$. Next,
similar to what's done in the related AoPS thread Number of solutions of two systems, we get
$$\begin{equation}\begin{aligned}
2\sum_{i=0}^{A}a_{i}(a_{i}-a_{i+1}) & = \color{blue}{\sum_{i=0}^{A}a_{i}^{2}} + \sum_{i=0}^{A}a_{i}^{2} - 2\sum_{i=0}^{A}a_{i}a_{i+1} \\
& = \color{blue}{a_{0}^{2} - a_{A+1}^2 + \sum_{i=1}^{A+1}a_{i}^{2}} + \sum_{i=0}^{A}a_{i}^{2} - \sum_{i=0}^{A}2a_{i}a_{i+1} \\
& = \color{blue}{a_{0}^{2} + (\sum_{i=0}^{A}a_{i+1}^{2}} - \sum_{i=0}^{A}2a_{i}a_{i+1} + \sum_{i=0}^{A}a_{i}^{2}) \\
& = a_{0}^{2} + \sum_{i=0}^{A}(a_{i+1}^{2} - 2a_{i}a_{i+1} + a_{i}^{2}) \\
& = a_{0}^{2} + \sum_{i=0}^{A}(a_{i+1} - a_{i})^{2} \\
& \gt 0
\end{aligned}\end{equation}$$
