한
Problem workshop

One great problem, three kinds of immersion

From a story to your own conjecture

Problem stories open curiosity. The problem workshop is the next place, where curiosity becomes a testable idea.

  1. 1 · DiscoverMeet the questionLearn when it appeared and why it still holds people’s attention.
  2. 2 · ChallengeTest one caseMake your first observation in five minutes with a drawing, calculation, or colors.
  3. 3 · DevelopYou are hereBuild on ideasPublish an observation and grow it through comparisons, conjectures, and counterexamples.
Week 12 · ClassicalOpen
Primary source

Brocard's Problem — When Is n!+1n! + 1 a Perfect Square?

Intermediate· Posed 1876A085692

Problem

Find all positive integers (n,m)(n, m) with n!+1=m2n! + 1 = m^2. Three known: (4,5),(5,11),(7,71)(4,5), (5,11), (7,71). Conjectured (Brocard 1876, Ramanujan 1913): no others — these triples are Brown numbers.

Why it matters

Posed by Henri Brocard (1876) and independently by Ramanujan (1913). Heuristic: n!n! contains so many small prime factors that n!+1n!+1 rarely matches a perfect-square divisor structure. Yet exactly n=4,5,7n = 4, 5, 7 hit — coincidence or pattern?

Progress so far

No other solutions for n≤109n \le 10^9 (Berndt & Galway 2000, since extended). If a strong form of the abc conjecture holds, only finitely many solutions exist (Overholt 1993). But abc itself remains open (Mochizuki's 2012 proposed proof is still under scrutiny).

Further reading

💡 Explore together, one line at a time(0 contributions)

Contributions are not ranked by popularity. Curator feedback names what is clear or reproducible, and peer signals mean someone understood or actually reproduced it.

What did you notice?

You do not need a complete proof. A small observation can open the next path.

Markdown + KaTeX supported (`$x^2$` inline, `$$\sum_{k=1}^n k$$` display)
0 / 3000 characters

moderation policy. Sign in after submitting if you want to edit or delete this attempt from another device.

Loading…

Related concepts