Detailed mathematical approach

Problem Summary

We count integer-sided triangles with one \(120^\circ\) angle opposite side \(c\), ordered so that \(a \le b \le c\), with the near-isosceles restriction \(b-a \le 100\) and the size bound \(c \le N\). The target value is \(T(10^{100})\), so a direct search over side lengths is hopeless. The solution must generate all valid families algebraically and then count only the distinct ordered triples.

Mathematical Approach

Once the angle is fixed at \(120^\circ\), the geometry turns into a quadratic Diophantine equation. The standard parameterization for integer \(120^\circ\) triangles then converts the near-isosceles condition into a Pell-type equation with a very small right-hand side.

Step 1: Turn the triangle condition into an arithmetic equation

By the law of cosines, a triangle with angle \(120^\circ\) opposite \(c\) satisfies

$c^2=a^2+b^2-2ab\cos 120^\circ=a^2+b^2+ab.$

So every admissible triangle corresponds to a positive integer solution of

$c^2=a^2+b^2+ab,\qquad a \le b \le c.$

This quadratic form is the norm form associated with Eisenstein integers, which is why a clean two-parameter description exists.

Step 2: Use the standard parameterization of primitive \(120^\circ\) triangles

For integers \(u>v>0\), define

$a^\ast=u^2-v^2,\qquad b^\ast=2uv+v^2,\qquad c^\ast=u^2+uv+v^2.$

A direct expansion shows that

$\left(c^\ast\right)^2=\left(a^\ast\right)^2+\left(b^\ast\right)^2+a^\ast b^\ast.$

Thus \((a^\ast,b^\ast,c^\ast)\) is a primitive integer \(120^\circ\) triangle, except that the two shorter sides are not automatically ordered. Every non-primitive solution is obtained by scaling:

$ (a,b,c)=\lambda\bigl(\min(a^\ast,b^\ast),\max(a^\ast,b^\ast),c^\ast\bigr),\qquad \lambda \ge 1.$

The counting problem is therefore reduced to generating primitive triangles and then determining how many scale factors \(\lambda\) remain legal.

Step 3: Convert the near-isosceles condition into a Pell-type equation

Write

$r=u-v>0,\qquad u=v+r.$

Then the difference between the two shorter sides becomes

$b^\ast-a^\ast=2uv+v^2-(u^2-v^2)=3v^2-r^2.$

Equivalently,

$r^2-3v^2=d,\qquad d=-(b^\ast-a^\ast).$

After ordering the shorter sides, their gap is exactly

$\bigl|\max(a^\ast,b^\ast)-\min(a^\ast,b^\ast)\bigr|=|d|.$

Scaling by \(\lambda\) multiplies every side difference by \(\lambda\), so the condition \(b-a \le 100\) becomes

$\lambda |d| \le 100.$

Therefore only nonzero integers \(d\) with \(|d|\le 100\) matter, and every primitive triangle for that \(d\) can be scaled only up to

$1 \le \lambda \le \left\lfloor\frac{100}{|d|}\right\rfloor.$

The case \(d=0\) would require \(r^2=3v^2\), impossible for positive integers, so the search really is over \(d\in\{-100,\dots,-1,1,\dots,100\}\).

Step 4: Solve the Pell-type equation orbit by orbit

For fixed \(d\), we must solve

$r^2-3v^2=d,\qquad r>0,\quad v>0.$

This is a Pell-type equation. Its positive solutions break into finitely many orbits under multiplication by the fundamental unit \(2+\sqrt{3}\). In coordinates, one forward step is

$r' = 2r+3v,\qquad v' = r+2v.$

If \((r,v)\) solves \(r^2-3v^2=d\), then so does \((r',v')\). The reverse unit \(2-\sqrt{3}\) gives

$r_- = 2r-3v,\qquad v_- = -r+2v,$

which preserves the same equation whenever both values stay positive. Repeating this reverse step until positivity would fail produces a reduced representative of the orbit. Starting from one reduced representative for each orbit and iterating forward generates every positive solution in that orbit.

Step 5: Reconstruct the triangle and count the valid scales

Once a solution \((r,v)\) is known, we recover

$u=v+r,$

$a^\ast=u^2-v^2,\qquad b^\ast=2uv+v^2,\qquad c^\ast=u^2+uv+v^2.$

The primitive triangle contributes all scaled copies with

$1 \le \lambda \le \min\left(\left\lfloor\frac{100}{|d|}\right\rfloor,\left\lfloor\frac{N}{c^\ast}\right\rfloor\right).$

Every such \(\lambda\) gives one admissible triangle after sorting the two shorter sides. Because repeated generation can occur, the implementation stores the final ordered triples in a set and counts each distinct triangle only once.

Worked Example

Take \(d=1\). Then

$r^2-3v^2=1$

has the positive solution \((r,v)=(2,1)\). Hence \(u=v+r=3\), so

$a^\ast=3^2-1^2=8,\qquad b^\ast=2\cdot 3\cdot 1+1^2=7,\qquad c^\ast=3^2+3\cdot 1+1^2=13.$

After ordering the shorter sides we obtain \((7,8,13)\), whose gap is indeed \(1\). For any bound \(N\ge 13\), the allowed scales are

$1 \le \lambda \le \min\left(100,\left\lfloor\frac{N}{13}\right\rfloor\right).$

Applying the forward recurrence once gives

$ (r',v')=(2\cdot 2+3\cdot 1,\;2+2\cdot 1)=(7,4).$

Now \(u=11\), which yields the primitive triangle \((104,105,181)\) after ordering. It again has short-side difference \(1\). This shows how one small value of \(d\) creates an infinite family, while the bound \(c\le N\) truncates the family to finitely many members.

How the Code Works

The C++, Python, and Java implementations all follow the same structure. They iterate through every signed difference parameter \(d\) with \(-100 \le d \le 100\) and \(d \ne 0\). For each \(d\), the maximum scale allowed by the near-isosceles restriction is precomputed as \(\lfloor 100/|d| \rfloor\).

Next, the implementation finds reduced positive solutions of the Pell-type equation \(r^2-3v^2=d\). Those reduced representatives are cached so that each orbit is discovered once. From every representative, the forward recurrence generated by \(2+\sqrt{3}\) is applied repeatedly. Each new pair \((r,v)\) produces one primitive triangle, and the orbit stops as soon as the corresponding \(c^\ast\) already exceeds \(N\).

For every primitive triangle, the two shorter sides are sorted, and every admissible scale factor

$1 \le \lambda \le \min\left(\left\lfloor\frac{100}{|d|}\right\rfloor,\left\lfloor\frac{N}{c^\ast}\right\rfloor\right)$

is tested. The resulting ordered triple is inserted into a set, which guarantees exact deduplication. Two useful checkpoints for this method are \(T(1000)=235\) and \(T(10^8)=1245\), after which the same machinery is applied to \(N=10^{100}\).

Complexity Analysis

The outer loop over \(d\) has constant size, since only \(200\) nonzero values are relevant. Along any fixed Pell orbit, the quantities \(r\), \(v\), and \(c^\ast\) grow roughly by the factor \(2+\sqrt{3}\), so the number of primitive triangles with \(c^\ast \le N\) is \(O(\log N)\) on that orbit. The number of scale factors per primitive triangle is at most \(100\), which is also a constant.

If \(M\) denotes the number of candidate triangles produced before deduplication, then generation is essentially linear in \(M\). Deduplication costs \(O(M\log M)\) in a tree-based set model, or near-linear expected time with hash-based sets. Memory usage is \(O(M)\). In practice this is tiny compared with any direct search over side lengths.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=582
2. Law of cosines: Wikipedia — Law of cosines
3. Pell's equation: Wikipedia — Pell's equation
4. Eisenstein integers: Wikipedia — Eisenstein integer
5. Diophantine equations: Wikipedia — Diophantine equation