How to use DeepSeek-Prover-V2 for free?
Deepseek has just dropped a new model, which is excellent in mathematical reasoning, that is DeepSeek Prover V2.
DeepSeek-Prover-V2 is an advanced language model specialized in formal theorem proving using the Lean 4 proof assistant.
https://medium.com/media/538c069770ed17513d674f770a75be3a/href
In simple terms,
DeepSeek-Prover-V2 is designed to support mathematicians and computer scientists in creating and validating formal proofs. Think of it as solving complex math problems with absolute logical precision. It leverages Lean 4, a specialized programming language for mathematical reasoning, to ensure every step of the proof is rigorously verified.
How to use DeepSeek-Prover-V2 for free?
https://medium.com/media/732eb9f0bb3e7b5c4d644ca2cbdf0c8b/href
OpenRouter is providing a free API key to run this new DeepSeek model. Yes, you heard it right. You can get the free API key for DeepSeek-Prover-v2 right now and use it to leverage for mathematical reasoning.
How to generate a free DeepSeek-Prover-V2 API?
- Visit OpenRouter
- Search for DeepSeek-Prover-V2 (free) & click
- Scroll below and generate a free API
- Use the code snippet below to run the new DeepSeek model
import requests
import json
response = requests.post(
url="https://openrouter.ai/api/v1/chat/completions",
headers={
"Authorization": "Bearer YOUR API",
"Content-Type": "application/json",
},
data=json.dumps({
"model": "deepseek/deepseek-prover-v2:free",
"messages": [
{
"role": "user",
"content": "Proof the pythogoras theorem"
}
],
})
)
print(response.content)
The output
{"id":"gen-1746019884-GH6HqDzRwnLG3GNzra6v","provider":"Chutes","model":"deepseek/deepseek-prover-v2:free","object":"chat.completion","created":1746019884,"choices":[{"logprobs":null,"finish_reason":"stop","native_finish_reason":"stop","index":0,"message":{"role":"assistant","content":"The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This can be written as: \n\n\\[ a^2 + b^2 = c^2 \\]\n\nwhere \\( c \\) is the length of the hypotenuse, and \\( a \\) and \\( b \\) are the lengths of the other two sides.\n\n### Proof Using Similar Triangles\n\n**Step 1: Construct a Right Triangle**\nConsider a right-angled triangle \\( ABC \\) with the right angle at \\( C \\). So, the hypotenuse is \\( AB = c \\), and the other two sides are \\( BC = a \\) and \\( AC = b \\).\n\n**Step 2: Draw the Altitude**\nDraw an altitude from \\( C \\) to the hypotenuse \\( AB \\), intersecting at \\( D \\). Let \\( AD = m \\) and \\( BD = n \\). Note that \\( m + n = c \\).\n\n\n\n**Step 3: Identify Similar Triangles**\nThe altitude divides the original triangle into two smaller triangles, \\( ACD \\) and \\( CBD \\), both of which are similar to the original triangle \\( ABC \\).\n\n1. \\( \\triangle ACD \\sim \\triangle ABC \\) because they both have a right angle and share \\( \\angle A \\). (AA similarity)\n - Therefore, \\( \\frac{AD}{AC} = \\frac{AC}{AB} \\Rightarrow \\frac{m}{b} = \\frac{b}{c} \\Rightarrow m = \\frac{b^2}{c} \\).\n\n2. \\( \\triangle CBD \\sim \\triangle ABC \\) because they both have a right angle and share \\( \\angle B \\). (AA similarity)\n - Therefore, \\( \\frac{BD}{BC} = \\frac{BC}{AB} \\Rightarrow \\frac{n}{a} = \\frac{a}{c} \\Rightarrow n = \\frac{a^2}{c} \\).\n\n**Step 4: Sum the Segments \\( m \\) and \\( n \\)**\nRecall that \\( m + n = c \\). Substituting the expressions from Step 3:\n\n\\[ \\frac{a^2}{c} + \\frac{b^2}{c} = c \\]\n\nMultiply both sides by \\( c \\):\n\n\\[ a^2 + b^2 = c^2 \\]\n\nThis completes the proof.\n\n### Alternate Proof Using Squares and Rearrangement\n\nThis is a visual proof often attributed to Indian mathematician Bhaskara, or sometimes to ancient Chinese mathematicians (\"Gougu theorem\").\n\n**Step 1: Construct Squares**\nConstruct a square with side length \\( a + b \\), and place four copies of the original right-angled triangle inside it. The triangles can be arranged in two ways to prove the theorem.\n\n**Arrangement 1:**\nIn the first arrangement, the four triangles are placed such that they form a square with side length \\( c \\) inside the larger square. The area of the larger square is \\( (a + b)^2 \\), and the area of the smaller square inside is \\( c^2 \\). The area of the four triangles is \\( 4 \\times \\frac{1}{2}ab = 2ab \\). Thus, the area of the larger square is the sum of the inner square and the four triangles:\n\n\\[ (a + b)^2 = c^2 + 2ab \\]\n\nExpanding \\( (a + b)^2 \\):\n\n\\[ a^2 + 2ab + b^2 = c^2 + 2ab \\]\n\nSubtract \\( 2ab \\) from both sides:\n\n\\[ a^2 + b^2 = c^2 \\]\n\n**Arrangement 2:**\nIn the second arrangement, the four triangles are placed such that they form two rectangles, each of size \\( a \\times b \\), inside the larger square. This leaves two smaller squares inside, one with area \\( a^2 \\) and the other with area \\( b^2 \\). Again, the area of the larger square is \\( (a + b)^2 \\), and the sum of the areas of the two smaller squares and the four triangles is:\n\n\\[ a^2 + b^2 + 2ab \\]\n\nSetting these equal:\n\n\\[ (a + b)^2 = a^2 + b^2 + 2ab \\]\n\nBut we already know that \\( (a + b)^2 = a^2 + 2ab + b^2 \\), so this is consistent. It also shows that rearranging the triangles in different ways can help visualize the relationship \\( a^2 + b^2 = c^2 \\).\n\nBoth of these proofs, using similar triangles and geometric arrangement, confirm the Pythagorean Theorem. The proof using similar triangles relies on algebraic manipulation, while the second proof is more visual and demonstrates the theorem by comparing areas.","refusal":null,"reasoning":null}}],"usage":{"prompt_tokens":10,"completion_tokens":1032,"total_tokens":1042,"prompt_tokens_details":null}}'
That’s it !!
You can even try the model for free on the HuggingFace inference section on the official page
deepseek-ai/DeepSeek-Prover-V2-671B · Hugging Face
Hope you try out the model
DeepSeek-Prover-V2 Free API was originally published in Data Science in Your Pocket on Medium, where people are continuing the conversation by highlighting and responding to this story.
