Picky Rectangles

Relevant wiki: Quadratic Diophantine Equations - Solve by Factoring

From this note , we know that the number of $x\times y$ rectangles in an $a\times b$ grid given that $x\leqslant a$ and $y\leqslant b$ is $(a-x+1)(b-y+1)$

Here, we have $a=150$ , $b=70$ , substituting them yields $(150-x+1)(70-y+1)=178 \\ (151-x)(71-y)=178$ Since $x$ and $y$ are integers, $x\leqslant151$ , $y\leqslant71$ and 178 can be factorised as $1\times178$ and $2\times89$ , we have $\begin{cases} 151-x=1\\71-y=178 \end{cases} \text{ or } \begin{cases} 151-x=2\\71-y=89 \end{cases} \text{ or } \begin{cases} 151-x=89\\71-y=2 \end{cases} \text{ or } \begin{cases} 151-x=178\\71-y=1 \end{cases}$ All of them results in $\begin{cases} x=150\\y=-107 \end{cases} \text{ or } \begin{cases} x=149\\y=-18 \end{cases} \text{ or } \begin{cases} x=62\\y=69 \end{cases} \text{ or } \begin{cases} x=-27\\y=70 \end{cases}$ Because $x$ and $y$ are both positive, hence $x=62$ , $y=69$ , $x+y=131$

178 = 2 * 89.
Answer.
= (150-89+1) + (70-2+1).
= 131