<s>[INST] If a rectangle can be split into n and into n+14 squares, then n=18. What is the meaning of this This problem seems to be a geometrical one, but actually it is an algebraic one.
If you split the rectangle into n squares and into n+k squares, then what does that means? What is this supposed to mean?
 [/INST] The rectangle area must add up exactly to the sums of the areas of the squares. Let $x$ be the side length of each of the $n$ squares and $y$ be the side length of each of the $n+k$ squares. Thus, you get the rectangle area is
$$nx^2 = (n+k)y^2 \tag{1}\label{eq1}$$
This can be adjusted to
$$\frac{n+k}{n} = \frac{x^2}{y^2} \tag{2}\label{eq2}$$
Thus, the left side must be the square of a rational number. In your case, $k = 14$. With $n = 18$, you get that $\frac{n+k}{n} = \frac{32}{18} = \frac{16}{9}$. This gives that $x = 4m$ and $y = 3m$ for some positive integer $m$. From \eqref{eq1}, the area of the rectangle would then be $18 \times (4m)^2 = 288m^2$.
You can do a similar type of check for other values of $k$. To determine a value of $n$ for any given value of $k$, consider $k = 2^a b$ where $a \ge 0$ and $b$ is an odd integer $\gt 1$. Then $n = 2^a\left(\frac{b-1}{2}\right)^2$ will give
\begin{align}
\frac{n+k}{n} & = \frac{2^a\left(\frac{b-1}{2}\right)^2 + 2^ab}{2^a\left(\frac{b-1}{2}\right)^2} \\
& = \frac{b^2 - 2b + 1 + 4b}{4\left(\frac{b-1}{2}\right)^2} \\
& = \frac{b^2 + 2b + 1}{(b-1)^2} \\
& = \frac{(b + 1)^2}{(b-1)^2} \tag{3}\label{eq3}
\end{align}
In the case with $k = 14 = 2 \times 7$, this gives that $b = 7$ and $n = 2 \times \left(\frac{6}{2}\right)^2 = 2 \times 9 = 18$ and \eqref{eq3} gives $\frac{8^2}{6^2} = \frac{4^2}{3^2} = \frac{16}{9}$ as noted above.
For any given $k$ which has a solution, there are at most a finite # of values of $n$ which work, with just the $1$ value of $n = 18$ for $k = 14$. To see this, note that $x$ must be greater than $y$ in \eqref{eq2}, so let $x = y + c$ for some positive integer $c$. Substituting this into \eqref{eq2}, expanding and then simplifying gives
$$\frac{n + k}{n} = \frac{y^2 + 2cy + c^2}{y^2} \\
1 + \frac{k}{n} = 1 + \frac{2cy + c^2}{y^2} \\
\frac{k}{n} = \frac{2cy + c^2}{y^2} \\
ky^2 = 2cny + c^2n \\
ky^2 - 2cny - c^2n = 0 \tag{4}\label{eq4}$$
This is a quadratic equation in $y$, so the quadratic formula gives
\begin{align}
y & = \frac{2cn \pm \sqrt{4c^2n^2 + 4c^2nk}}{2k} \\
& = \frac{cn \pm c\sqrt{n^2 + nk}}{k} \tag{5}\label{eq5}
\end{align}
For $y$ to be an integer requires the part in the square roots in \eqref{eq5} to be a perfect square, i.e.,
$$n^2 + nk = (n + d)^2 \tag{6}\label{eq6}$$
for some positive integer $d$. However, since $n^2 \lt n^2 + 2nk \lt \left(n  + \frac{k}{2}\right)^2$, then
$$0 \lt d \lt \frac{k}{2} \tag{7}\label{eq7}$$
Next, expanding the right hand side of \eqref{eq6}, moving the $n$ term over gives
$$n(k - 2d) = d^2 \tag{8}\label{eq8}$$
This requires that $d^2$ be an integral multiple of $k - 2d$. For $k = 14$, using \eqref{eq7}, this requires checking $d$ from $1$ to $6$ for any which also satisfy \eqref{eq8}. In this case, $d = 1$ gives $12n = 1$, $d = 2$ gives $10n = 4$, $d = 3$ gives $8n = 9$, $d = 4$ gives $6n = 16$, $d = 5$ gives $4n = 25$, so only $d = 6$ works to get that $n(2) = 36 \implies n = 18$ is the one and only solution.
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