Perfect Square AP Ratio

We might as well assume that $a,b,c$ are pairwise coprime, since any common factor cancels out when evaluating $\frac{b-a}{c-b}$ . Since $a^2 + c^2 = 2b^2$ , both $a$ and $c$ must be odd, and so we can write $a = u-v$ , $c = u+v$ where $u,v$ are coprime positive integers with $v < u$ , in which case $u^2 + v^2 = b^2$ . There are now two cases:

• We can write $u = m^2 - n^2$ , $v = 2mn$ , $b = m^2 + n^2$ , which makes $\frac{b-a}{c-b} = \frac{m+n}{m-n} = 1 + 2\big(\tfrac{m}{n} - 1\big)^{-1}$ . Since we must have $u > v$ , we must have $m^2 -2mn - n^2 > 0$ , and hence $\tfrac{m}{n} > 1 + \sqrt{2}$ . Thus $\frac{b-a}{c-b} < 1 + \sqrt{2}$ , and we can approach $1 + \sqrt{2}$ as closely as we like by choosing $m,n$ so that $\tfrac{m}{n}$ is a close to $1 + \sqrt{2}$ as we like from above.

• We can write $u = 2mn$ , $v = m^2 - n^2$ , $b = m^2 + n^2$ , which makes $\frac{b-a}{c-b} = \frac{m}{n}$ . .Since we must have $u > v$ , we must have $n^2 - 2mn - m^2 > 0$ , so that $\tfrac{m}{n} < 1 + \sqrt{2}$ . Thus $\frac{b-a}{c-b} < 1 + \sqrt{2}$ , and we can again get as close to $1 + \sqrt{2}$ as we like by letting $\tfrac{m}{n}$ get as close to $1 + \sqrt{2}$ as we like from underneath.

Nice problem Calvin, I missed 2root2 and therefore missed by like millionth

Very nicely done!

For completeness, we should explain how to acheive $a \rightarrow 0, c \rightarrow \sqrt{2} b$ . In fact, that stictly speaking cannot be done as $a$ is a positive integer. Actually, what we want is $x = \frac{a}{b} \rightarrow 0 , y = \frac{ c}{ b} \rightarrow \sqrt{2}$ .