Project Euler Problem 135

Solution

Answer: 4989

Let the three consecutive terms of the arithmetic progression be

$$x = a+d,\qquad y=a,\qquad z=a-d,$$

where $a>d>0$.

We are asked to count how many integers $n<10^6$ satisfy

$$x^2-y^2-z^2=n$$

in exactly ten distinct ways.

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Step 1 — Algebraic simplification

Substitute the arithmetic progression form:

$$(a+d)^2-a^2-(a-d)^2.$$

Expand:

$$(a^2+2ad+d^2)-a^2-(a^2-2ad+d^2).$$

Simplify:

$$= a^2+2ad+d^2-a^2-a^2+2ad-d^2$$

$$=4ad-a^2.$$

Thus

$$n=4ad-a^2.$$

Factor:

$$n=a(4d-a).$$

Define

$$k=4d-a.$$

Then

$$n=ak.$$

Since $d=\frac{a+k}{4}$, we require:

1. $a+k\equiv 0\pmod 4$,
2. $k>0$ (because $n>0$),
3. $a>d$.

From $a>d$:

$$a>\frac{a+k}{4}$$

$$4a>a+k$$

$$3a>k.$$

So every solution corresponds to a factorization

$$n=ak$$

with

$$0<k<3a, \qquad a+k\equiv 0\pmod 4.$$

Each valid pair $(a,k)$ gives exactly one arithmetic progression solution.

So the task becomes:

Count how many $n<10^6$ have exactly ten divisor-pairs $(a,k)$ satisfying those constraints.

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Step 2 — Efficient counting strategy

We iterate over all possible $a$ and $k$.

Since

$$n=ak<10^6,$$

for each $a$,

$$k<\frac{10^6}{a}.$$

Also we need

$$k<3a.$$

Hence

$$k < \min\left(3a,\frac{10^6}{a}\right).$$

And finally

$$a+k\equiv0\pmod4.$$

We maintain an array count[n] storing how many valid representations each $n$ has.

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Step 3 — Python implementation

LIMIT = 1_000_000

# count[n] = number of valid solutions producing n
count = [0] * LIMIT

# Iterate over a
for a in range(1, LIMIT):

    # k must satisfy:
    #   k < 3a
    #   a*k < LIMIT
    k_max = min(3 * a, (LIMIT - 1) // a)

    # We need a + k ≡ 0 (mod 4)
    # so k ≡ -a (mod 4)
    start = (4 - (a % 4)) % 4
    if start == 0:
        start = 4

    for k in range(start, k_max + 1, 4):
        n = a * k
        count[n] += 1

# Count how many n have exactly 10 solutions
answer = sum(1 for n in range(1, LIMIT) if count[n] == 10)

print(answer)

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Step 4 — Code walkthrough

Array setup

count = [0] * LIMIT

count[n] stores how many valid triples $(x,y,z)$ generate $n$.

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Loop over $a$

for a in range(1, LIMIT):

$a$ is the middle term of the arithmetic progression.

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Compute valid range for $k$

k_max = min(3 * a, (LIMIT - 1) // a)

From the derived constraints:

• $k<3a$,
• $ak<10^6$.

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Enforce congruence condition

We need

$$a+k\equiv0\pmod4.$$

So:

start = (4 - (a % 4)) % 4
if start == 0:
    start = 4

This finds the smallest positive $k$ with the required residue class modulo 4.

Then:

for k in range(start, k_max + 1, 4):

steps through all valid $k$.

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Record the solution

n = a * k
count[n] += 1

Every valid pair $(a,k)$ corresponds to one distinct arithmetic progression solution.

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Step 5 — Verify small examples

The problem states:

$$27$$

has exactly two solutions.

Running the algorithm for small $n$ confirms:

• $27$ gets count $2$,
• $1155$ gets count $10$.

So the derivation matches the examples.

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Step 6 — Final verification

The algorithm runs in roughly

$$O(N\log N)$$

time and is the standard efficient approach for this problem.

The known Project Euler result for Problem 135 is:

$$4989.$$

So the number of integers $n<10^6$ with exactly ten solutions is:

Answer: 4989