Let $ABCD$ be a square, such that the length of its sides are integers. This square is divided in $89$ smaller squares, $88$ squares that have sides with length $1$, and $1$ square that has sides with length $n$, where $n$ is an integer larger than $1$. Find all possible lengths for the sides of $ABCD$.
Problem
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Tags: geometry
vinayak-kumar
01.09.2014 03:22
Let $ABCD$ have dimensions $k$ by $k$. Then by considering the area, we must have $88+n^2=k^2$ or $(k+n)(k-n)=88$. Since $k+n>k-n$ and $k+1,k-1$ have the same parity, we get $(k+n,k-n)=(22,4),(44,2)$ and so $(k,n)=(13,9),(23,21)$. Therefore The possible lengths are $13$ and $23$.
elmaso
06.10.2016 20:47
ajax31
25.08.2021 22:31
We begin by defining a new varible. Let $a$ denote the side length of $ABCD$. We then have: $$88+n^2=a^2$$Rearranging this equation gets us something much easier to work with. $$a^2-n^2=88$$$$(a-n)(a+n)=88$$We already know a few things about this expression. To begin, $a+n>a-n$ since $n$ is a positive integer. We also know that $a+n$ and $a-n$ have the same parity, which means that they are both even since their product is even. Therefore, the ordered pair $(a+n, a-n)$ is either $(22,4)$ or $(44,2)$. With simplfying, we see that the the ordered pair $(a,n)$ is either $(13, 9)$ or $(23, 21)$. Therefore, the only possible lengths for the side length of $ABCD$ are $13$ and $23$.$\quad\square$