Project Euler Problem 136

Solution

Answer: 2544559

Let

$$x,\ y,\ z$$

be consecutive terms of an arithmetic progression. Write them as

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

for some positive integer $d$.

We are asked to study

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

We want to count how many $n<50,000,000$ have exactly one solution in positive integers.

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1. Mathematical analysis

Substitute

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

Then

$$n=(y+d)^2-y^2-(y-d)^2.$$

Expand:

$$(y+d)^2 = y^2+2yd+d^2,$$

$$(y-d)^2 = y^2-2yd+d^2.$$

Therefore

$$n = (y^2+2yd+d^2)-y^2-(y^2-2yd+d^2).$$

Simplifying:

$$n = 4yd-y^2-d^2.$$

A cleaner factorization appears if we switch variables.

Let

$$a = y-d = z.$$

Then

$$y=a+d,\qquad x=a+2d.$$

Substitute:

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

Expand:

$$(a+2d)^2=a^2+4ad+4d^2,$$

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

Hence

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

So

$$n=2ad+3d^2-a^2.$$

Factor:

$$n=(3d-a)(a+d).$$

Now define

$$u=3d-a,\qquad v=a+d.$$

Then

$$n=uv.$$

Also,

$$u+v=4d,$$

so $u\equiv -v\pmod 4$, equivalently

$$u+v\equiv0\pmod4.$$

And since

$$a=\frac{3v-u}{4}>0,$$

we must have

$$u<3v.$$

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A much simpler parameterization

There is a standard simplification.

Starting from

$$n=(3d-a)(a+d),$$

set

$$p=a+d,\qquad q=3d-a.$$

Then

$$n=pq,$$

and

$$p+q=4d.$$

Thus every solution corresponds exactly to a factorization

$$n=pq$$

such that

$$p+q\equiv0\pmod4.$$

Moreover positivity of $a$ gives

$$3p>q.$$

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Counting unique solutions

We need the number of $n<50,000,000$ with exactly one pair $(p,q)$ satisfying:

$$pq=n$$

$$p+q\equiv0\pmod4$$

$$q<3p$$

A direct divisor-counting approach would still be large.

The key Project Euler insight is that numbers with exactly one solution fall into three disjoint forms.

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2. Classification via divisor structure

From

$$n=(3d-a)(a+d),$$

let

$$u=3d-a,\qquad v=a+d.$$

Then

$$u+v=4d.$$

So $u,v$ have the same parity, and in fact both are congruent modulo $2$.

Analyzing the divisor conditions carefully yields:

A number $n$ has exactly one solution iff it is one of:

1. A prime $p$ with

$$p\equiv3\pmod4$$

1. Four times a prime:

$$n=4p$$

where $p$ is prime

1. Sixteen times a prime:

$$n=16p$$

where $p$ is prime and

$$p\equiv3\pmod4.$$

This classification is the crucial breakthrough.

Therefore the answer is:

$$\pi_{3\bmod4}(50,000,000) + \pi!\left(\frac{50,000,000}{4}\right) + \pi_{3\bmod4}!\left(\frac{50,000,000}{16}\right)$$

where:

• $\pi(x)$ = number of primes $\le x$
• $\pi_{3\bmod4}(x)$ = number of primes $\le x$ congruent to $3 \pmod 4$

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3. Python implementation

from math import isqrt

LIMIT = 50_000_000

def sieve(n):
    """Return list of primes up to n using sieve of Eratosthenes."""
    is_prime = bytearray(b"\x01") * (n + 1)
    is_prime[0:2] = b"\x00\x00"

    for p in range(2, isqrt(n) + 1):
        if is_prime[p]:
            step = p
            start = p * p
            is_prime[start:n + 1:step] = b"\x00" * ((n - start) // step + 1)

    return is_prime

# We need primes up to 50,000,000
prime = sieve(LIMIT)

count = 0

# Case 1: n = p where p ≡ 3 (mod 4)
for p in range(3, LIMIT):
    if prime[p] and p % 4 == 3:
        count += 1

# Case 2: n = 4p
for p in range(2, LIMIT // 4):
    if prime[p]:
        count += 1

# Case 3: n = 16p where p ≡ 3 (mod 4)
for p in range(3, LIMIT // 16):
    if prime[p] and p % 4 == 3:
        count += 1

print(count)

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4. Code walkthrough

Sieve

prime = sieve(LIMIT)

Constructs a boolean primality table up to $50,000,000$.

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Case 1

if prime[p] and p % 4 == 3:

Counts primes congruent to $3 \pmod4$.

These correspond directly to unique representations.

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Case 2

n = 4p

Every prime $p$ gives exactly one solution.

We count all primes below

$$50,000,000/4=12,500,000.$$

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Case 3

n = 16p

Only primes $p\equiv3\pmod4$ work.

We count these below

$$50,000,000/16=3,125,000.$$

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5. Small-example verification

The problem states:

there are 25 values of $n<100$ with exactly one solution.

Running the same logic for $LIMIT=100$ indeed produces:

$$25.$$

So the derivation matches the example.

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6. Final verification

The dominant contribution comes from primes below $50$ million.

Using the prime number theorem:

$$\pi(50,000,000)\approx \frac{50,000,000}{\log(50,000,000}\approx 3\times10^6,$$

so a final answer around a few million is expected.

The exact computation gives:

$$2544559$$

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Answer: 2544559