The largest ball I can fold from 1856 sheets

I had 1856 sheets of paper and a simple-sounding question: what’s the biggest round ball I can fold, and what would another 1000 sheets buy me? It became the most satisfying wrong-turn-then-right-turn of the whole thread.

The setup is $N = 30T \le \text{budget}$, so for 1856 sheets, $T \le 61$, i.e. $b^2 + bc + c^2 \le 61$. My instinct was calculus: treat it as an optimization, differentiate, find the critical points. That instinct produces real math, a genuinely pretty page of it, and the wrong answer.

The trap is that the curve $b^2 + bc + c^2 = 61$ isn’t a constraint you optimize $T$ over. It is a level set of $T$. Every point on it has $T = 61$ by definition, so there is nothing to maximize along it.¹1 Implicit differentiation gives $dc/db = -(2b+c)/(b+2c)$, which is $0$ at $c = -2b$ and vertical at $c = -b/2$. Those are the tilted ellipse’s extreme points, real geometry, just unrelated to counting units. I was trying to maximize the very quantity that draws the curve.

The real problem is discrete. I don’t want a point on the ellipse; I want the integer lattice point inside it with the largest $T$. That’s a number-theory question: find the largest value that $b^2 + bc + c^2$ actually achieves (a Loeschian number) at or below $\lfloor \text{budget}/30 \rfloor$, then read off its $(b,c)$. There’s no clean formula, because Loeschian membership depends on prime factorization, but the search is only a couple of steps:

import math

def representation(T):
    for c in range(math.isqrt(T) + 1):
        disc = 4*T - 3*c*c
        if disc < 0:
            break
        s = math.isqrt(disc)
        if s*s == disc and (s - c) % 2 == 0 and (s - c) // 2 >= c:
            return ((s - c) // 2, c)
    return None

def largest_ball(sheets):
    T = sheets // 30
    while representation(T) is None:   # walk down to the nearest Loeschian number
        T -= 1
    b, c = representation(T)
    return 30 * T, (b, c), T

Run it. With 1856 sheets, $T$ caps at 61, and 61 is Loeschian on the nose, it’s prime and $\equiv 1 \pmod 3$. That gives $(b,c) = (5,4)$, an 1830-unit ball, 26 sheets to spare. With another 1000 (2856 total), $T$ caps at 95, but $95 = 5 \cdot 19$ fails and 94 fails too, so it drops to $93 = 3 \cdot 31$, giving $(b,c) = (7,4)$, a 2790-unit ball, $\{3,5+\}(7,4)$.

Two things I didn’t expect. Both winners are Class III chiral ($b \neq c$, both nonzero), so the largest ball I can fold is a giant left- or right-handed spiral, 1830 parallelograms all leaning the same way like wheat in wind. And the extra 1000 sheets only moved $T$ from 61 to 93, because the paper budget grows linearly while $T$ grows like the square of the ball’s frequency. Doubling how round the ball looks costs four times the paper. You pay quadratically for roundness. Roundness is a luxury good.

The lesson that stuck: the continuous-optimization reflex was right about the method and wrong about the object. The moment the variable becomes “how many integer lattice points fit,” calculus hands the problem to number theory.