A rectangle can be divided into $n$ equal squares. If the same rectangle can also be divided into $n+ 76$ equal squares then find $n$.
I tried with taking dimensions of squares as $x$ and $y$. Then I made the respective lenghts and breadths equal for two cases. Also, I equated the combined areas, but then I am not able to find the solutions.
Let $w$ and $h$ be the width and height of the rectangle in arbitrary unit. WOLOG, we will assume $w \ge h$.
In order for the rectangle to be covered perfectly by $n$ squares. $w$ and $h$ need to be commensurable. i.e. $\displaystyle\;\frac{w}{h} \in \mathbb{Q}$. Choose a new unit so that $\displaystyle\;w, h \in \mathbb{Z}_{+}$ and $\gcd(w,h) = 1$.
If we can cover the rectangle by $n$ squares of side $s$, then both $\frac{w}{s}$ and $\frac{h}{s}$ have to integers. Furthermore, since $\left.\frac{w}{s}\right/\frac{h}{s} = \frac{w}{h}$ and $\gcd(w,h) = 1$, there is a $p \in \mathbb{Z}_{+}$ so that $\frac{w}{s} = pw$ and $\frac{h}{s} = ph$. This implies $n = p^2 wh$.
By a similar argument, there is a $q \in \mathbb{Z}_{+}$ so that $n + 76 = q^2 wh$. This leads to
$$(q+p)(q-p)wh = 76 = 2^2\times 19\tag{*1}$$
Notice $q-p$ and $q+p$ has same parity, there are two possibilities:
Both $q-p$ and $q+p$ are odd
In this case, $wh$ contains the $2^2$ factor from $76$. Since $19$ is a prime and $\gcd(w,h) = 1$, we find
$$(q+p, q-p, wh ) = (19,1,4) \implies (q,p,w,h) = (10,9,4,1)\\ \implies n = 9^2\times 4\times 1 = 324$$
Both $q-p$ and $q+p$ are even
We can rewrite $(*1)$ as $$\frac{q+p}{2}\frac{q-p}{2} w h = 19$$ Since $\frac{q+p}{2} > \frac{q-p}{2}$ are both positive integers, $\frac{q+p}{2}\frac{q-p}{2} > 1$. Since $19$ is a prime, this leads to $$\left(\frac{q+p}{2},\frac{q-p}{2}, wh \right) = (19,1,1) \implies (q,p,w,h) = (20,18,1,1)\\ \implies n = 18^2 \times 1^2 = 324$$
Combine these, we can conclude $n = 324$ is the unique solution of this problem.
The number $76$ equals $4 * 19$. Now as well-known the difference between two successive squares goes as $1, 3, 5, 7, 9...$ Based on this series we can easily establish that $19= 100 - 81 = 10^2 - 9^2$. The multiplicative factor $4$ can be taken into account without difficulty, since it equals $2$ squared. This way we arrive at the important intermediate result that the number $76$ can be expressed as the difference between two squares as follows:
$$76 = 400 - 324 = 20^2 - 18^2$$
Now if we take $324$ squares of length $10$, or $400$ squares of length $9$, both combinations have the same surface area ($= 32400$). From this we conclude that $N = 324$ is the solution.
Furthermore it follows that the rectangle has length $360$ and width $90$ (unless a square with sides $180$ is also allowed).
