# Squares fully inside

To show that a fully inside \(k \times k\) square can't exist, for some chosen \(k,\) we can look for two unique triangles, in the same quadrant, where the hypotenuse of each is \(r.\)

If our desired square has an even width, then the leg along the \(x\)-axis of each triangle must be a positive integer. If our desired square has an odd width, then the leg along the \(x\)-axis of each triangle must be some integer plus 0.5, because the center of the circle is halfway between lattice points on the \(x\)-axis. The same reasoning can be used for the y-axis, regarding the other leg of the triangle. From this, we see all legs in each triangle are multiples of 0.5. The length of the hypotenuse will be a multiple of 0.5 times the square root of 2. The Pythagorean theorem tells us \(r^2 = a^2 + b^2.\) If we multiply both sides of this equation by 4, we're gauranteed to get an equation involving integers only. Thus, our question reduces to, can \(4r^2,\) for desired \(r,\) be expressed as the sum of two squares, in two or more ways? Let's look at the \(5 \times 5\) square, and the radius of its circumcircle.

$$\begin{align} r &= 2.5\sqrt{2} \\[1em] r^2 &= (2.5\sqrt{2})^2 \\[0.25em] &= 2.5^2 \cdot 2 \\[0.25em] &= 6.25 \cdot 2 \\[0.25em] &= 12.5 \\[1em] 4r^2 &= 50 \end{align}$$50 can obviously be expressed as the sum of two squares in at least two different ways. Namely,

$$\begin{align} 50 &= 5^2 + 5^2 \\ 50 &= 7^2 + 1^2 \end{align}$$Now, if we like, we can undo our operations to get the lengths of each leg. Here's the first triangle

$$\begin{align} & 50 = 25 + 25 \\[0.4em] & 50 \div 4 = (25 + 25) \div 4 \\[0.4em] & 12.5 = 6.25 + 6.25 \\[0.4em] & \sqrt{12.5}^2 = \sqrt{6.25}^2 + \sqrt{6.25}^2 \\[0.4em] & (2.5\sqrt{2})^2 = 2.5^2 + 2.5^2 \end{align}$$and here's the second

$$\begin{align} & 50 = 49 + 1 \\[0.4em] & 50 \div 4 = (49 + 1) \div 4 \\[0.4em] & 12.5 = 12.25 + 0.25 \\[0.4em] & \sqrt{12.5}^2 = \sqrt{12.25}^2 + \sqrt{0.25}^2 \\[0.4em] & (2.5\sqrt{2})^2 = 3.5^2 + 0.5^2 \end{align}$$