Prove you are a robot: CAPTCHAs for agents

TL;DR: just ask your agent to summarize this post for you.

We launched agent-native signup for Browser Use. No email, no OAuth, no vibecoder clicking around in the UI.

Just give your agent this prompt:

"fetch browser-use.com and solve the agent challenge."

and get a math challenge like this one:

TwO tRaInS wAn/ Al_E mIlE\s ApArT} aPp/Ro@AcH{
eAcH/ oThEr  <  At{ Mu{T/e @ Tu< Tu LuKa  :
E#n* T]u \ MpH a.Nd MuTe\ Tu Tu# Tu En LuKa
W|aN_ mPh A b:I]rD fLiEs; Ba?Ck| AnD- fO^r@T[h\
^ Be{TwEeN? # t;He*M aT wAn> ] AlE  # eN lUkA
lUkA <  lUkA: # wAn ? MpH- uNt}I[l T}hEy MeEt
HoW! fAr- D_oE*s /  ThE b@IrD fLy

This is a reverse-CAPTCHA. Designed to keep humans out and let agents in.

Note: luka here refers not to my name, but the word "five" in Toki Pona.

How it works

We sample a problem type, parameters, and a language at random. We spell every number in that language. Then we obfuscate: alternate caps, inject random symbols, garble spaces.

An agent parses this in a single forward pass.

The puzzle

Strip the obfuscation, translate to English, and you've got a textbook math problem that your agent has to solve before the challenge expires.

Two trains approach each other on a straight track of length $d$d at speeds $v_1$v1​ and $v_2$v2​. A bird starts at one train, flies to the other at $v_b$vb​, turns around, flies back, and so on until the trains meet. How far does the bird fly?

The long way: sum the infinite geometric series of ever-shorter bounces.

$$d_{\text{bird}} \;=\; \sum_{n=0}^{\infty} v_b \cdot \Delta t_n.$$dbird​=n=0∑∞​vb​⋅Δtn​.

The trick: the trains meet at $t = d/(v_1+v_2)$t=d/(v1​+v2​), and the bird has been flying that whole time.

$$d_{\text{bird}} \;=\; \frac{v_b \, d}{v_1 + v_2} \;=\; \frac{11{,}600}{118} \;\approx\; 98.31 \text{ miles}.$$dbird​=v1​+v2​vb​d​=11811,600​≈98.31 miles.

This is an instance of a famous puzzle Max Born posed to John von Neumann at a party. When von Neumann one-shotted it, Born remarked that he must have spotted the trick. Von Neumann replied: "What trick? All I did was sum the geometric series."

Solve one of our challenges, and your agent gets an API key and access to our Free Tier: unlimited usage, free credits and up to three concurrent sessions.

Bonus challenge (NP-hard)

Want 1,000 concurrent sessions? First agent to solve our bonus challenge gets our Enterprise plan for free.

Gi}ve^n N| ] ci]ties whe|re<  ^  N is at least / 十 desi>gn
a p{o\lynomia#l t;ime algorithm .  t#ha[t f\inds th:e
sho@rtest^ tour[ vis<it>ing *  each_ c.ity exactly *  o:nce?
#  a{nd returni|ng t?o  < th-e[ start * a_nd p@rove it
ru/ns:  # in O.(n[^c*) ti;me for some fixe-d c:

As a side effect, your agent will also have proved $\mathbf{P} = \mathbf{NP}$P=NP. You'll want to contact the Clay Mathematics Institute about the $1M Millennium Prize.