Monday, Jan 23rd

Problem
A rectangle can be divided into $n$ equal squares and also into $n+98$ equal squares. If the area is $n$, find its sides.

Solution
Let $x$ be the side of  each of the $n$ equal squares the rectangle can be divided into. Likewise define $y$ to be when it is divided into $n+98$ squares. Then  $nx^2=(n+98)y^2=n$ and $x=1$ and $y=\sqrt{\frac{n}{n+98}}$. Since there must be an integer number of $y$ squares on each row (column) and also and integer number of $x$ rectangles in each row (column), then $y$ has to be a rational number.

If $\text{gcd}(n,n+98)=1$, then $y^2$ is an irreducible fraction and both $n$ and $n+98$ have to be squares. Since $98\equiv 2 \text{ mod }4$, this is impossible.

Therefore $\text{gcd}(n,n+98)>1$, and hence $\text{gcd}(n,98)>1$. Let $n=dm$ where $d=\text{gcd}(n,98)$. Then $y^2=\frac{m}{m+98/d}$ and $\text{gcd}(m,m+98/d)=1$. Thus $m$ and $m+98/d$ have to be squares and that happens if and only if $2|d$. Let $m=p^2$ and $m+98/d=q$. If $d=2$, $49=(q-p)(q+p)$ and the only solutions are $q=25$ and $p=24$. If $d=14$, $7=(q-p)(q+p)$ with solutions $q=4$ and $p=3$.  If $d=98$, there are no solutions. Hence the possibilities for $n$ are $n=2\cdot 24^2$ and $n=14\cdot 3^2$.