Match Stick Squares

For a square of side length $17$ , we can first calculate the total number of match sticks used:

For the vertical grids, we need $17\times (17+1$ = 306) match sticks. Similarly, the horizontal grids will also yield the same amount.

Therefore, there are totally $612$ match sticks used.

Suppose then that the new rectangle has side of $a$ and $b$ . The total amount of match sticks used can be evaluated as:

$612 = b(a+1) + a(b+1) = 2ab + a + b$

Then let's generalize the formula for square of side $x$ and let $a = x - c$ and $b = x + d$ for some positive integers $c$ and $d$ .

Hence, $2x(x+1) = 2(x-c)(x+d) + (x-c) + (x+d)$ .

$0 = 2(d-c)x - 2cd +(d-c)$

$2x = \dfrac{2cd}{d-c} -1$

Substitute $x = 17$ , we will get: $35(d-c) = 2cd$ .

Since $35$ is odd, $d-c$ is even.

Then let $d-c = 2k$ for some integer $k$ .

$35(2k) = 2c(c+2k)$

$35k = c(c+2k)$

$0 = c^2 + (2k)c -35k$

Solving for $c$ , we will get: $c = \dfrac{-2k \pm \sqrt{4k^2 + 4\cdot 35k}}{2} = -k \pm \sqrt{k^2 + 35k}$ .

Since $c$ is an integer, the square root residue must also be an integer.

In order to maximize the area, the side difference must be as minimal as possible. Then $k = 1$ applies.

Therefore, $c = 5$ and $d = 7$ : $35\times 2 = 2\times 5\times 7$ .

As a result, $a = x-5 = 17-5 = 12$ and $b = x+7 = 17+7 = 24$ .

Checking the answer: $612 = 2ab + a + b = 2\times 12\times 24 + 12 +24 = 576 + 36$ , which is correct.

As a result, the square will have an area of $17\times 17 = 289$ , and the rectangle will have an area of $12\times 24 = 288$ .

Thus, the area difference is $289-288 = \boxed{1}$ .