Solution

Answer: 756872327473

We seek integer solutions to the condition

$$P(\text{BB})=\frac12$$

where a box contains $b$ blue discs and $r$ red discs, with total

$$n=b+r.$$

The probability of drawing two blue discs without replacement is

$$\frac{b}{n}\cdot\frac{b-1}{n-1}=\frac12.$$

So we must solve

$$\frac{b(b-1)}{n(n-1)}=\frac12.$$

Equivalently,

$$2b(b-1)=n(n-1).$$

We are asked for the first solution with

$$n>10^{12}.$$

Step 1 — Transforming the equation

Start from

$$2b(b-1)=n(n-1).$$

Expand:

$$2b^2-2b=n^2-n.$$

Multiply by $4$:

$$8b^2-8b=4n^2-4n.$$

Now complete squares:

$$8b^2-8b+2=4n^2-4n+1.$$

Thus

$$2(2b-1)^2=(2n-1)^2+1.$$

Rearrange:

$$(2n-1)^2-2(2b-1)^2=-1.$$

Define

$$x=2n-1,\qquad y=2b-1.$$

Then we obtain the Pell equation

$$x^2-2y^2=-1.$$

Step 2 — Pell equation recurrence

The negative Pell equation

$$x^2-2y^2=-1$$

has infinitely many solutions generated recursively.

The first few are:

$$(1,1),\ (7,5),\ (41,29),\ (239,169),\dots$$

These correspond to:

$$n=\frac{x+1}{2},\qquad b=\frac{y+1}{2}.$$

For example:

$x=239,\ y=169$

gives

$$n=120,\qquad b=85,$$

which matches the problem statement.

The standard recurrence for solutions is

$$x_{k+1}=3x_k+4y_k,$$

$$y_{k+1}=2x_k+3y_k.$$

Converting back to $(n,b)$, we get:

$$n_{k+1}=3n_k+4b_k-3,$$

$$b_{k+1}=2n_k+3b_k-2.$$

Starting from the known solution

$$(n,b)=(21,15),$$

we repeatedly generate larger solutions until $n>10^{12}$.

Step 3 — Python implementation

# Project Euler Problem 100

LIMIT = 10**12

# Initial known solution:
# total discs = 21
# blue discs  = 15
n = 21
b = 15

# Generate larger solutions using the Pell recurrence
while n <= LIMIT:
    n, b = (
        3 * n + 4 * b - 3,
        2 * n + 3 * b - 2
    )

print(b)

Step 4 — Code walkthrough

LIMIT = 10**12

We need the first arrangement with more than one trillion total discs.

n = 21
b = 15

Start from the first nontrivial solution given in the problem.

while n <= LIMIT:

Keep generating solutions until the total exceeds $10^{12}$.

n, b = (
    3 * n + 4 * b - 3,
    2 * n + 3 * b - 2
)

This is the Pell-equation recurrence transformed back into $(n,b)$-coordinates.

Each iteration produces the next valid arrangement.

print(b)

Output the number of blue discs in the first valid arrangement exceeding the limit.

Step 5 — Checking small cases

Starting from $(n,b)=(21,15)$:

Next recurrence gives:

$$n=120,\qquad b=85.$$

Check:

$$\frac{85}{120}\cdot\frac{84}{119} = \frac{7140}{14280} = \frac12.$$

Correct.

Continuing the recurrence eventually yields:

$$n=1,070,379,110,497,$$

$$b=756,872,327,473.$$

Indeed,

$$n > 10^{12}.$$

Final verification

The solution satisfies the Pell equation structure.
The recurrence generates only exact integer solutions.
The resulting total exceeds $10^{12}$.
The blue count is slightly smaller than the total, as expected.

Answer: 756872327473