A number theory problem by Priyanshu Mishra

This is probably not the simplest way to do this, but here is what I did:

There is a rational number $k>0$ such that $k=\sqrt{(2 a - b)/(2 a + b)}$ . Let $k = x/y$ for integers $x > 0, y > 0, gcd(x,y)=1$ . So $a=b (1+k^2)/(2(1-k^2))$ , and $p = b k / 4$ . $b k = b x / y$ is an integer, so $b$ is divisible by $y$ , say $b = b_1 y$ , for integer $b_1$ , i.e., $4 p = b_1 x$ . Thus, (1) $p | b_1$ or (2) $p | x$ .

Case 1: $x = p x_1$ , so $4 = b_1 x_1$ , and we have three cases $(b_1, x_1) = \{(1,4),(2,2),(4,1)\}$ . Tracking back, we three cases for $a,b$ in terms of $p, y$ .

$\text{}\left[\left\{\left\{a\to \frac{y \left(16 p^2+y^2\right)}{2 \left(y^2-16 p^2\right)},b\to y\right\},\left\{a\to \frac{y \left(4 p^2+y^2\right)}{y^2-4 p^2},b\to 2 y\right\},\left\{a\to \frac{2 y \left(p^2+y^2\right)}{y^2-p^2},b\to 4 y\right\}\right\}\right]$

Case 2: $b_1 = p b_2$ , so $4 = b_2 x$ , and we have three cases $(b_2, x) = \{(1,4),(2,2),(4,1)\}$ . Tracking back, we three cases for $a,b$ in terms of $p, y$ .

$\text{}\left[\left\{\left\{a\to \frac{p y \left(y^2+16\right)}{2 \left(y^2-16\right)},b\to p y\right\},\left\{a\to \frac{p y \left(y^2+4\right)}{y^2-4},b\to 2 p y\right\},\left\{a\to \frac{2 p y \left(y^2+1\right)}{y^2-1},b\to 4 p y\right\}\right\}\right]$

So we want the largest prime $p$ such that there exists an integer $y>0$ so that at least one of the six expressions above for $a$ is a positive integer. Note that each of these six cases is covered by the single case

$\text{}\left[a=\frac{\text{y} \left(4 p^2+\text{y}^2\right)}{\text{y}^2-4 p^2}\right], y>2 p$

(Reduce each of the six cases to this by considering y as even or a multiple of $p$ ).

Let $u = -2 p + y, v = 2 p + y$ . Then we have $(u+v)^3/(4 u v), v - u = 4 p, u > 0$ , which implies $(u^3 + v^3)/(u v)$ is an integer. Thus, $u | v^3 = (u + 4p)^3$ , so $u | 2^6 p^3$ . Thus, $v | u^3 = (v - 4p)^3$ , so $v | 2^6 p^3$ . There are 28 factors of $2^6 p^3$ , which we can list out. Then, we consider all 378 (28 choose 2) pairs of such factors and solve for $p$ such that the difference between these factors is $4 p$ . We see that the possible values of $p$ are 2,3,5,7,17. We then check these with

$\text{}\left[a=\frac{\text{y} \left(4 p^2+\text{y}^2\right)}{\text{y}^2-4 p^2}\right], y>2 p$

and see that the largest prime that works is 5, for example take $a=39, b=30$ . I used Mathematica to handle the 378 cases.