Megan is jealous of her older sister's $a\times b$ inch Zac Efron poster, so she cuts out a rectangle around his face, which has dimensions $\frac{a}{2}-2$ by $\frac{b}{3}-5$, and steals it. Megan's older sister decides she's over Zac Efron anyways, and uses the rest of the paper as wrapping paper. She wraps an $11$-inch cube, wasting no paper, and has $3$ square inches leftover. If $a$ and $b$ are integers and $b>a$, and the difference between the length and width of the poster is less than $50$, find the dimensions of the poster.
Express your answer as the ordered pair $(a,b)$.
I don't really know how to start this problem. Any solutions would be awesome.
Hint:
The key to such tasks is to extract the mathematical information:
The poster is a rectangle with sides $a$ and $b$. And implicitly: $a, b > 0$.
The cut part is a rectangle with sides $a/2 -2$ and $b/3-5$.
$$ A_{\text{poster}} - A_{\text{cut}} = A_{\text{cube}} + 3 \, \text{inch}^2 \\ a_{\text{cube}} = 11 \, \text{inch} $$
$$ a, b \in \mathbb{Z} \\ a < b $$
$$ \lvert a - b \rvert < 50 $$
We can easily calculate the three areas and end up with a conic equation of the type $$ A x^2 + B x y + C y^2 + D x + E y = F $$ and the inequalities, $$ a, b > 0 \\ a < b \\ b - a < 50 $$ Due to the integer constraints there should be a finite set of solutions.
Hint:
If there is no easier way, the title seems a bit suspicious, the systematic approach seems to transform (link, link) the quadratic form to a Pell-type equation (link) $$ X^2 - D Y^2 = N $$ which can be solved by continued fractions, if I remember correctly. And then weed out some solutions by those inequalities.
Update:
Yes, the title was suspicious. See this video (link).
So SFFT reminds me somewhat of "completing the square": \begin{align} 5 ab + 15 a + 4b &= 4434 \iff \\ 5a (b+3) + 4b &= 4434 \iff \\ 5a (b+3) + 4b + 4\cdot3 &= 4434 + 4\cdot 3 \iff \\ 5a (b+3) + 4(b+3) &= 4446 \iff \\ (5a+4)(b+3) &= 2 \cdot 3^2 \cdot 13 \cdot 19 \end{align}
This will now give the opportunity to assign the $N=5$ factors to each of the $M=2$ factors in parentheses on the left hand side: \begin{array}{cc|cc|cc|c} 5a+4 & b+3 & 5a+4 & b+3 & a & b & \text{constraints OK} \\ \hline 2 & 3 \cdot 3 \cdot 13 \cdot 19 & 2 & 2223 & ! & & \\ 3 & 2 \cdot 3 \cdot 13 \cdot 19 & 3 & 1482 & ! & & \\ 13 & 2 \cdot 3 \cdot 3 \cdot 19 & 13 & 342 & ! & & \\ 19 & 2 \cdot 3 \cdot 3 \cdot 13 & 19 & 234 & 3 & 231 & \text{no}, \Delta < 50 \\ \hline 2 \cdot 3 & 3 \cdot 13 \cdot 19 & 6 & 741 & ! & & \\ 2 \cdot 13 & 3 \cdot 3 \cdot 19 & 26 & 171 & ! & & \\ 2 \cdot 19 & 3 \cdot 3 \cdot 13 & 38 & 117 & ! & & \\ 3 \cdot 3 & 2 \cdot 13 \cdot 19 & 9 & 494 & 1 & 491 & \text{no}, \Delta < 50 \\ 3 \cdot 13 & 2 \cdot 3 \cdot 19 & 39 & 114 & 7 & 111 & \text{no}, \Delta < 50 \\ 3 \cdot 19 & 2 \cdot 3 \cdot 13 & 57 & 78 & ! & & \\ 13 \cdot 19 & 2 \cdot 3 \cdot 3 & 247 & 18 & ! & & \\ \hline 3 \cdot 13 \cdot 19 & 2 \cdot 3 & 741 & 6 & ! & & \\ 3 \cdot 3 \cdot 19 & 2 \cdot 13 & 171 & 26 & ! & & \\ 3 \cdot 3 \cdot 13 & 2 \cdot 19 & 117 & 38 & ! & & \\ 2 \cdot 13 \cdot 19 & 3 \cdot 3 & 494 & 9 & 98 & 6 & \text{no}, a < b \\ 2 \cdot 3 \cdot 19 & 3 \cdot 13 & 114 & 39 & 22 & 36 & \text{yes} \\ 2 \cdot 3 \cdot 13 & 3 \cdot 19 & 78 & 57 & ! & & \\ 2 \cdot 3 \cdot 3 & 13 \cdot 19 & 18 & 247 & ! & & \\ \hline 3 \cdot 3 \cdot 13 \cdot 19 & 2 & 2223 & 2 & & ! & \\ 2 \cdot 3 \cdot 13 \cdot 19 & 3 & 1482 & 3 & & ! & \\ 2 \cdot 3 \cdot 3 \cdot 19 & 13 & 342 & 13 & ! & & \\ 2 \cdot 3 \cdot 3 \cdot 13 & 19 & 234 & 19 & 46 & 16 & \text{no}, a < b \\ \end{array}