# 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}