"bad primes", and prime factorizations

First, by using "bad primes", and prime factorizations, etc.) decide if you can make squares with the following areas on a geoboard (with the usual geoboard set up - points one unit apart in a square area, and with the four corners of the square lying on grid points). Note, please don't actually draw out such a square, just explain whether or not it's possible, and feel free to use a calculator to factor the numbers. You visit Francis Su's neat Math Fun Facts website: https://www.math.hmc.edu/funfacts/ffiles/20008.5.s
a) 1105 (try a factor of 13 to start with)
b) 630
c) 882
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2) Next, recall the neat geoboard/unit square grid paper result The "GC" (Group Conjecture!) - that the area of a polygon drawn on unit square grid paper (i.e. so that the polygon's vertices are all on lattice points of the grid) is equal to I + B/2 - 1, where I is the number of points in the interior of the polygon, and B is the number of points on the boundary of the polygon. The question is whether it's possible to prove that the formula works for all polygons. One way to go about this is to start by looking at whether it works for rectangles and triangles and then build up from there, given that every polygon can be created by attaching
triangles and rectangles together (actually it can all be done with just triangles). For this first homework problem prove that the Group Conjecture works for a rectangle that's length X by Y (where X and Y are integers, and the rectangle is lined up squarely with the grid points - i.e. not tilted). The area of such a rectangle is clearly X times Y, now does the GC correctly predict that? Note that to show this, you can't
just give one example, but you need to show that it works for a general X by Y rectangle - i.e. you need to figure out how many points are on the boundary of an X by Y rectangle along with how many points are in the interior... then check the GC formula...
3) Next verify that the formula works for right triangles. To simplify your calculations, assume that the triangle's hypotenuse doesn't go through any grid points (note that this doesn't limit the result, as you could always break a larger right triangle into smaller right triangles if the hypotenuse of the larger triangle did go through some grid points). Note that once again you need to prove this for general right triangles (i.e. it's not enough to just show that it works for just one specific example - assume that you've got a triangle with shorter sides A and B, for example).4) The final step in the process of proving that the formula is valid for all possible polygons will be to show that if you break a larger polygon into a series of rectangles and triangles, that the formula still holds. To do this you'll need to show that if the formula works for two polygons, P1 and P2 (as we know it does for rectangles and right triangles, for example), then it will work a polygon, P, that's created by merging the two polygons. To check this, figure out what would happen referencing the following polygon merger example . Start by assuming that the area of P1 is given by I1+ B1/2 - 1 where I1 and B1 are the number of interior and boundary points, respectively, for polygon P1 (and likewise that the area of P2 is given by I2 + B2/2 - 1). Now show that that the polygon P given as the combination of P1 and P2 must also have its area given by the I + B/2 -1 formula, by figuring out what I and B are for the combination polygon P,
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