Scaling Open + Decentralized AI
What’s Next
Join Us
Authors
Vincent
Johannes
Announcements
February 28, 2025

$15M to Build a Peer-to-Peer AI Protocol

Prime Intellect is building a peer-to-peer protocol for compute and intelligence, enabling collective creation, ownership, and access to sovereign open-source AI. We’re moving beyond centralized AI to empower anyone—from solo GPU operators to global datacenters—to contribute compute, code, or capital and shape the open and decentralized AI ecosystem.

Today, we’re announcing that we’ve raised $15M in funding, led by Founders Fund, with participation from Menlo Ventures, Andrej Karpathy (EurekaAI, Tesla, OpenAI), Clem Delangue (Hugging Face), Dylan Patel (SemiAnalysis), Tri Dao (Together.AI), Balaji Srinivasan (Network School), and Emad Mostaque (Stability AI, Intelligent Internet), Jake Medwell (8VC Co-Founder), Brendan McCord (Cosmos Institute), Sandeep Nailwal (Polygon) and many others

This brings our total funding to over $20M, accelerating our work on peer to peer compute and intelligence.

Scaling Open + Decentralized AI

Prime Intellect is already operating at scale.

  • Compute: Our Compute Exchange, connects individual developers, startups, cloud providers, and datacenters, creating a decentralized GPU marketplace that makes AI compute cheaper and more accessible.
  • Decentralized Model Training: We successfully trained a 10B-parameter model decentralized across three continents with our PRIME framework , proving that decentralized training can match centralized efficiency—without a single point of control.
  • Open-Source AI Research:
    • INTELLECT-1, the first 10B-parameter model trained on decentralized infrastructure.
    • METAGENE-1, a bio-foundation model for early pandemic detection and global health applications.
    • INTELLECT-MATH, a frontier model for math reasoning trained with RL.
    • GENESYS + SYNTHETIC-1, a distributed RL effort generating the largest synthetic reasoning dataset from DeepSeek-R1.
  • Verifiable Compute: We’ve published TOPLOC, a locality sensitive hashing scheme for efficient verifiable inference.
  • Peer-to-peer compute and intelligence network: The Prime Intellect protocol is live on testnet, enabling a trustless, community-driven ecosystem where contributors benefit from the models they help create.

What’s Next

We’re building an AI future that is open, permissionless, and collectively owned.

  • Expand Compute Markets – Scaling our marketplace so anyone can rent or sell GPUs on demand.
  • Intelligence Markets – Enabling a peer to peer market of all different AI models, and enabling anyone to compete against the incumbents to serve AI models cheaply.
  • Scale Decentralized Training – Integrating TOPLOC-style verification mechanisms for inference and training to scale decentralized training, synthetic data generation, distributed reinforcement learning and inference to frontier models
  • Enable Collective Ownership –Extending our design to support the financialization of models, agents, datasets and code to enable novel revenue streams for open source AI
  • Launching collaborations with other leading projects in open source and decentralized AI to bootstrap an ecosystem of researchers and developers collectively advancing open research
  • And much more…

Join Us

If you’re excited to help shape the future of open-source AI, we’d love to hear from you and invite researchers, engineers, and visionaries worldwide to help build this decentralized infrastructure:

  • Contribute: Provide compute, code, or capital to our peer to peer AI protocol. Get Started
  • Collaborate: We’re hiring engineers and researchers at the intersection of AI, distributed systems and protocol development Careers
  • Connect: Follow @primeintellect on Twitter/X or join our Discord community.

A future of truly open superintelligence demands shared effort. Let’s build it together.

‍

Eurus-2-PRIME
System Prompt
When tackling complex reasoning tasks, you have access to the following actions. Use them as needed to progress through your thought process. [ASSESS] [ADVANCE] [VERIFY] [SIMPLIFY] [SYNTHESIZE] [PIVOT] [OUTPUT] You should strictly follow the format below: [ACTION NAME] # Your action step 1 # Your action step 2 # Your action step 3 ... Next action: [NEXT ACTION NAME]
User
Quadratic polynomials P(x) and Q(x) have leading coefficients 2 and −2, respectively. The graphs of both polynomials pass through the two points (16,54) and (20,53). Find P(0)+Q(0).
Eurus-2-PRIME
[ASSESS]

We are given two quadratic polynomials P(x) and Q(x) with leading coefficients 2 and −2, respectively. The graphs of both polynomials pass through the points (16,54) and (20,53). We need to find the value of P(0)+Q(0).

[ADVANCE]

Let's express the quadratic polynomials P(x) and Q(x) in the general form:
P(x)=2x2+bx+c
Q(x)=−2x2+dx+e

Since the graphs of both polynomials pass through the points (16,54) and (20,53), we can substitute these points into the equations of P(x) and Q(x) to form a system of equations.

For P(x):
P(16)=2(16)2+16b+c=54
512+16b+c=54
16b+c=−458(Equation 1)

P(20)=2(20)2+20b+c=53
800+20b+c=53
20b+c=−747(Equation 2)

For Q(x):
Q(16)=−2(16)2+16d+e=54
−512+16d+e=54
16d+e=566(Equation 3)

Q(20)=−2(20)2+20d+e=53
−800+20d+e=53
20d+e=853(Equation 4)

[ADVANCE]

Now, we can solve the system of equations for P(x):
Subtract Equation 1 from Equation 2:
(20b+c)−(16b+c)=−747−(−458)
4b=−289
b=−2894

Substitute b=−2894 into Equation 1:
16(−2894)+c=−458
−1156+c=−458
c=698

So, for P(x), we have:
P(x)=2x2−2894x+698
P(0)=698

Next, we can solve the system of equations for Q(x):
Subtract Equation 3 from Equation 4:
(20d+e)−(16d+e)=853−566
4d=287
d=2874

Substitute d=2874 into Equation 3:
16(2874)+e=566
1148+e=566
e=−582

So, for Q(x), we have:
Q(x)=−2x2+2874x−582
Q(0)=−582

[ADVANCE]

Finally, we can find P(0)+Q(0):
P(0)+Q(0)=698+(−582)=116

So, the value of P(0)+Q(0) is 116.
INTELLECT-MATH-7B-SFT (ours)
System Prompt
Solve the following math problem efficiently and clearly. Think carefully and step by step about your response and reason before providing a final response. Conclude your response with: \n\nTherefore, the final answer is: boxedanswer. I hope it is correct.\n\nWhere [answer] is just the final number or expression that solves the problem.
User
Quadratic polynomials P(x) and Q(x) have leading coefficients 2 and −2, respectively. The graphs of both polynomials pass through the two points (16,54) and (20,53). Find P(0)+Q(0).
INTELLECT-MATH-7B-SFT
So I've got this problem here about quadratic polynomials P(x) and Q(x). P(x) has a leading coefficient of 2, and Q(x) has a leading coefficient of −2. Both polynomials pass through the points (16,54) and (20,53). I need to find P(0)+Q(0).

First, since both P(x) and Q(x) are quadratic polynomials, I can express them in the general form:

P(x)=2x2+bx+c

Q(x)=−2x2+dx+e

Wait, actually, the leading coefficient is the coefficient of the highest power of x, which for a quadratic is x2. So yes, P(x)=2x2+bx+c and Q(x)=−2x2+dx+e.

Now, both polynomials pass through the points (16,54) and (20,53). That means when x=16, P(16)=54, and when x=20, P(20)=53. Similarly for Q(x).

So, I can set up equations using these points.

First, for P(x):

P(16)=2(16)2+b16+c=54

Similarly, P(20)=2(20)2+b20+c=53

Let me calculate 162 and 202:

162=256

202=400

So,

P(16):2256+16b+c=54⇒512+16b+c=54⇒16b+c=54−512⇒16b+c=−458

P(20):2400+20b+c=53⇒800+20b+c=53⇒20b+c=53−800⇒20b+c=−747

Now, I have two equations:

1) 16b+c=−458

2) 20b+c=−747

I can subtract equation 1 from equation 2 to eliminate c:

(20b+c)−(16b+c)=−747−(−458)

20b+c−16b−c=−747+458

4b=−289

So, b=−289/4=−72.25

Wait, but coefficients are usually integers, but maybe not. Let me check.

Now, plug b back into one of the equations to find c. Let's use equation 1:

16(−72.25)+c=−458

−1156+c=−458

c=−458+1156=698

So, P(x)=2x2−72.25x+698

Hmm, P(0) would be c, which is 698.

Now, let's do the same for Q(x).

Q(16)=−2∗(16)2+d16+e=54⇒−2256+16d+e=54⇒−512+16d+e=54⇒16d+e=54+512⇒16d+e=566

Q(20)=−2∗(20)2+d20+e=53⇒−2400+20d+e=53⇒−800+20d+e=53⇒20d+e=53+800⇒20d+e=853

Now, equations for Q(x):

1) 16d+e=566

2) 20d+e=853

Subtract equation 1 from equation 2:

(20d+e)−(16d+e)=853−566

4d=287

d=287/4=71.75

Now, plug d back into equation 1:

1671.75+e=566⇒1148+e=566⇒e=566−1148=−582

So, Q(x)=−2x2+71.75x−582

Therefore, Q(0)=e=−582

Now, P(0)+Q(0)=698+(−582)=116

Wait, but the problem says "present the answer in LaTeX format: \boxed{Your answer}"

So, I should write 116

But let me double-check my calculations to make sure I didn't make any mistakes.

First, for P(x):

P(16)=2(256)+16b+c=512+16b+c=54⇒16b+c=−458

P(20)=2∗(400)+20b+c=800+20b+c=53⇒20b+c=−747

Subtracting: 4b=−289⇒b=−72.25

Then, c=−458−16∗(−72.25)=−458+1156=698

Similarly for Q(x):

Q(16)=−2∗(256)+16d+e=−512+16d+e=54⇒16d+e=566

Q(20)=−2∗(400)+20d+e=−800+20d+e=53⇒20d+e=853

Subtracting: 4d=287⇒d=71.75

Then, e=566−16∗71.75=566−1148=−582

So, P(0)=698, Q(0)=−582, sum is 116.

Seems correct.

Final Answer

116
INTELLECT-MATH-7B (ours)
System Prompt
Solve the following math problem efficiently and clearly. Think carefully and step by step about your response and reason before providing a final response. Conclude your response with: \n\nTherefore, the final answer is: boxedanswer. I hope it is correct.\n\nWhere [answer] is just the final number or expression that solves the problem.
User
Quadratic polynomials P(x) and Q(x) have leading coefficients 2 and −2, respectively. The graphs of both polynomials pass through the two points (16,54) and (20,53). Find P(0)+Q(0).
INTELLECT-MATH-7B
So I have two quadratic polynomials, P(x) and Q(x), with leading coefficients 2 and -2, respectively. Both pass through the points (16,54) and (20,53). I need to find P(0)+Q(0).

First, since both are quadratic polynomials, I can express them in the general form:

P(x)=2x²+bx+c

Q(x)=−2x²+dx+e

But since both pass through the points (16,54) and (20,53), I can set up equations using these points.

For P(x):

P(16)=2∗(16)²+b⋅16+c=54

P(20)=2(20)²+b⋅20+c=53

Similarly for Q(x):

Q(16)=−2(16)²+d⋅16+e=54

Q(20)=−2(20)²+d⋅20+e=53

Let's compute the squares:

16²=256

20²=400

So for P(x):

2256+16b+c=54⇒512+16b+c=54⇒16b+c=54−512=−458...(1)

2400+20b+c=53⇒800+20b+c=53⇒20b+c=53−800=−747...(2)

For Q(x):

−2256+16d+e=54⇒−512+16d+e=54⇒16d+e=54+512=566...(3)

−2400+20d+e=53⇒−800+20d+e=53⇒20d+e=53+800=853...(4)

Now, I have two systems of linear equations:

For P(x):

16b+c=−458...(1)

20b+c=−747...(2)

Subtract equation (1) from equation (2):

(20b+c)−(16b+c)=−747−(−458)

4b=−289⇒b=−2894

Now plug b back into equation (1):

16(−2894)+c=−458⇒−1156+c=−458⇒c=−458+1156=698

So P(x)=2x²−2894x+698

For Q(x):

16d+e=566...(3)

20d+e=853...(4)

Subtract equation (3) from equation (4):

(20d+e)−(16d+e)=853−566

4d=287⇒d=2874

Now plug d back into equation (3):

16(2874)+e=566⇒1148+e=566⇒e=566−1148=−582

So Q(x)=−2x2+2874x−582.

Now, P(0)=c=698.

Q(0)=e=−582.

Therefore, P(0)+Q(0)=698+(−582)=116.

Final Answer

116.
Authors
Vincent
Johannes
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Acknowledgements

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