简单测试了一下deepseek的初等数学能力

[Veggiebed]

比如这个简单题目 x^2 + x + 1 = 0 ( mod 11) 的整数解是啥？deepseek转了半天，说server busy，please try later。

又改成问这个题目的解有几个（因为无解，所以等于是提示DS了）？ 转了半天，还是server busy，please try later。

是真的用的人太多了？

前面问了它一个级数收敛的问题，它立即就答了，答得不错。

按说这个初等数论题也最多是同样难度的，看来LLM还是这个样子，没刷过的题，就不会做。

[irisyuan]

被ddos攻击了，从前天开始基本就没法用了。

[Veggiebed]

被ddos攻击了，从前天开始基本就没法用了。

但是问他个简单级数收敛题，它就立即给出答案。看来主要还是靠刷题。

[irisyuan]

你再问他同样的问题，他不一定能答出来，你多试试就知道了，同样问题问10遍，起码9次他会说server busy。就是忙不过来了

[mmking]

我帮你问了一下，看来这个AI看人下菜碟啊

你再问他同样的问题，他不一定能答出来，你多试试就知道了，同样问题问10遍，起码9次他会说server busy。就是忙不过来了

[Veggiebed]

你再问他同样的问题，他不一定能答出来，你多试试就知道了，同样问题问10遍，起码9次他会说server busy。就是忙不过来了

可能是用的人太多，问了一个问题两次，一次能答，一次只是不停地转， 然后server busy。

[Veggiebed]

完了，现在拒绝回答了，问啥都是server busy

[Zephyrca]

比如这个简单题目 x^2 + x + 1 = 0 ( mod 11) 的整数解是啥？deepseek转了半天，说server busy，please try later。

又改成问这个题目的解有几个（因为无解，所以等于是提示DS了）？ 转了半天，还是server busy，please try later。

是真的用的人太多了？

前面问了它一个级数收敛的问题，它立即就答了，答得不错。

按说这个初等数论题也最多是同样难度的，看来LLM还是这个样子，没刷过的题，就不会做。

我觉得这个deepseak是不是靠数据造假出来结果的？非常值得怀疑

[irisyuan]

完了，现在拒绝回答了，问啥都是server busy

从前天就开始了。我估计都不一定就是ddos，国内那么多有需求又没办法access ChatGPT的用户，纯纯就是用到瘫了

[Veggiebed]

从前天就开始了。我估计都不一定就是ddos，国内那么多有需求又没办法access ChatGPT的用户，纯纯就是用到瘫了

应该不是attack，就是用到瘫痪了

[Veggiebed]

这回又能用了，有时会做错，比如这个明显答案的问题：x^2 = 10 (mod 30), 居然用了Chinese remainder theorem后说无解。

[（ヅ）]

半个小时我的本地deepseek-r1:32b 终于出完结果嘞


>>> integer solution of x^2 + x + 1 = 0 ( mod 11)
<think>
Alright, so I need to find all integer solutions for the equation \(x^2 + x + 1 = 0 \mod{11}\). Hmm, okay. Let me
think about how to approach this.

First off, solving quadratic equations modulo a prime number like 11 can sometimes be done by trying out all
possible residues from 0 up to 10 and seeing which ones satisfy the equation. Since 11 is not too large, that
might be manageable. But before jumping into that, maybe there's a smarter way.

I remember something about quadratic equations modulo primes. There's this thing called the discriminant. For a
general quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\). If the discriminant is a
square modulo the prime, then there are solutions; otherwise, there aren't.

So in this case, let's compute the discriminant. Here, \(a = 1\), \(b = 1\), and \(c = 1\), so:

\(D = (1)^2 - 4(1)(1) = 1 - 4 = -3\).

Hmm, okay, so we have a discriminant of \(-3\) modulo 11. I need to check if \(-3\) is a quadratic residue modulo
11. In other words, does there exist some number \(y\) such that \(y^2 \equiv -3 \mod{11}\)?

Let me compute \(-3\) mod 11 first. Since 11 - 3 = 8, so \(-3 \equiv 8 \mod{11}\). So I need to check if 8 is a
quadratic residue modulo 11.

To do that, one method is to use Euler's criterion, which says that for an odd prime \(p\), and integer \(a\) not
divisible by \(p\), \(a^{(p-1)/2} \equiv 1 \mod{p}\) if \(a\) is a quadratic residue, and \(-1\) otherwise.

So let me compute \(8^{5} \mod{11}\). Why 5? Because (11 - 1)/2 = 5. Let's calculate:

\(8^1 = 8\)

\(8^2 = 64\), which is 64 - 5*11=64-55=9 mod 11.

\(8^3 = 8 * 9 = 72\). 72 divided by 11: 6*11=66, so 72-66=6. So \(8^3 \equiv 6 \mod{11}\).

\(8^4 = 8 * 6 = 48\), which is 48 - 4*11=48-44=4 mod 11.

\(8^5 = 8 * 4 = 32\). 32 - 2*11=32-22=10 mod 11. So \(8^5 \equiv 10 \mod{11}\).

Since 10 is not congruent to 1, Euler's criterion tells me that 8 is not a quadratic residue modulo 11. Therefore,
the discriminant isn't a square in this field, which suggests that there are no solutions.

Wait, but let me double-check because sometimes even if I think I'm correct, it's better to be thorough. Maybe
trying all possible residues from 0 to 10 will confirm whether any satisfy the equation \(x^2 + x + 1 \equiv 0
\mod{11}\).

Alright, let me create a table of values for each x in [0, 10] and compute \(x^2 + x + 1\) mod 11.

Starting with x=0:

\(0^2 + 0 + 1 = 1\). 1 mod 11 is 1. Not zero.

x=1:

\(1 + 1 + 1 = 3\). 3 mod 11 is 3. Nope.

x=2:

4 + 2 + 1 =7. 7 mod 11 is 7. Still not.

x=3:

9 + 3 + 1=13. 13-11=2. Not zero.

x=4:

16 + 4 + 1=21. 21 - 2*11=21-22=-1≡10 mod11. Still not zero.

x=5:

25 +5 +1=31. 31 -2*11=9. Not zero.

x=6:

36 +6 +1=43. 43 - 3*11=43-33=10. Not zero.

x=7:

49 +7 +1=57. 57 - 5*11=57-55=2. Nope.

x=8:

64 +8 +1=73. 73 -6*11=73-66=7. Not zero.

x=9:

81 +9 +1=91. 91 -8*11=91-88=3. Nope.

x=10:

100 +10 +1=111. 111 -10*11=111-110=1. Not zero.

Hmm, so none of the residues from 0 to 10 satisfy the equation \(x^2 + x + 1 \equiv 0 \mod{11}\). So that confirms
it: there are no solutions modulo 11 for this quadratic equation.

Wait a second, but I just remembered something. Quadratic equations over finite fields can have 0, 1, or 2
solutions. Since the discriminant wasn't a square, we have zero solutions here, which matches with our exhaustive
check above.

Alternatively, maybe there's another method involving completing the square? Let me try that for practice.

Starting with \(x^2 + x + 1 \equiv 0 \mod{11}\).

Completing the square:

\(x^2 + x = -1 \mod{11}\)

Take half of the coefficient of x, which is 1/2. Since we're working mod11, 1/2 is equivalent to 6 because
2*6=12≡1 mod11.

So, \( (x + 6)^2 - 36 ≡ -1 \mod{11}\).

Wait, let me compute that step by step:

\(x^2 + x = (x + 1/2)^2 - (1/2)^2\). So,

\( (x + 6)^2 - 36 ≡ -1 \mod{11} \)

Compute \( -36 mod11 \): 36 /11 is 3*11=33, so 36-33=3. Thus, -36≡-3≡8 mod11.

So:

\( (x + 6)^2 ≡ -1 + 36 mod11\), which simplifies to \( (x + 6)^2 ≡ (-1) + (36 mod11) \).

Wait, hold on. Let me redo that step because I might have messed up.

We had:

\( x^2 + x = -1 \mod{11} \)

Complete the square:

\( x^2 + x + (1/4) = -1 + (1/4) \mod{11} \)

So,

\( (x + 1/2)^2 ≡ (-1) + (1/4) mod11 \)

Compute \(1/4\) modulo11: since 4*3=12≡1 mod11, so 1/4≡3 mod11.

Thus:

Left side: \( (x + 6)^2 \mod{11} \) because 1/2 is 6 in mod11.

Right side: (-1) + 3 ≡ 2 mod11.

So the equation becomes:

\( (x + 6)^2 ≡ 2 mod11 \)

Now, so we need to solve \( y^2 ≡ 2 mod11 \), where y = x +6. So does 2 have square roots modulo11?

Let me check which squares give 2 mod11.

Compute squares of numbers from 0 to 10:

0²=0

1²=1

2²=4

3²=9

4²=16≡5

5²=25≡3

6²=36≡3

7²=49≡5

8²=64≡9

9²=81≡4

10²=100≡1.

So the squares modulo11 are 0,1,3,4,5,9. So 2 isn't a square modulo11 either. Therefore, equation \( y^2 ≡2
\mod{11}\) has no solution, so original equation also has no solutions.

Wait, but earlier when computing Euler's criterion for discriminant 8 (from the first method), we saw that 8 was
not a quadratic residue as well? Or wait, in the second method, we arrived at needing to solve y²≡2. So yeah, same
result: no solution.

So all methods converge on there being no solutions to this equation in integers mod11.

Therefore, I think it's settled. There are no integer solutions modulo11 for \( x^2 +x +1 ≡0 \mod{11} \).

**Final Answer**
There are no integer solutions. The final answer is \boxed{\text{No solution}}.
</think>

To determine the integer solutions of the quadratic equation \(x^2 + x + 1 \equiv 0 \pmod{11}\), we can use
several methods:

1. **Exhaustive Check**:
- We check all residues from 0 to 10 modulo 11.
- None of these values satisfy the equation \(x^2 + x + 1 \equiv 0 \pmod{11}\).

2. **Quadratic Residue Check**:
- The discriminant of the quadratic equation is calculated as \(\Delta = b^2 - 4ac = 1 - 4 = -3\).
- Converting \(-3\) modulo 11, we get \(8\). We check if 8 is a quadratic residue modulo 11.
- The squares modulo 11 are calculated: 0, 1, 3, 4, 5, 9. Since 8 is not among these, it is not a quadratic
residue.

3. **Completing the Square**:
- Rewrite the equation as \(x^2 + x = -1 \pmod{11}\).
- Complete the square: \((x + 6)^2 - 36 \equiv -1 \pmod{11}\).
- Simplifying, we get \((x + 6)^2 \equiv 2 \pmod{11}\). Checking if 2 is a quadratic residue modulo 11, we find
it is not.

Since all methods confirm that there are no solutions, we conclude:

\[
\boxed{\text{No solution}}
\]

[TheMatrix]

我帮你问了一下，看来这个AI看人下菜碟啊

烈害。

[闪光的二猫]

我发现不开deep search比较蠢。昨天看网上有人黑 说不知道3.11 和3.9 哪个大。我立刻试了一下果然如此 而且重复问每次回答不一样 有的时候说3.11 大给出一大堆推理过程，说 所以 结论是 3.9 大…

推理过程说的也离谱 说3.11 等于 3.10

打开deep search 以后几次还可以

[irisyuan]

我发现不开deep search比较蠢。昨天看网上有人黑 说不知道3.11 和3.9 哪个大。我立刻试了一下果然如此 而且重复问每次回答不一样 有的时候说3.11 大给出一大堆推理过程，说 所以 结论是 3.9 大…

推理过程说的也离谱 说3.11 等于 3.10

打开deep search 以后几次还可以

不开deepthink就没意义了，就好比普通chatgpt和o1的差别，天差地别

[Maja101]

半个小时我的本地deepseek-r1:32b 终于出完结果嘞


>>> integer solution of x^2 + x + 1 = 0 ( mod 11)
<think>
Alright, so I need to find all integer solutions for the equation \(x^2 + x + 1 = 0 \mod{11}\). Hmm, okay. Let me
think about how to approach this.

First off, solving quadratic equations modulo a prime number like 11 can sometimes be done by trying out all
possible residues from 0 up to 10 and seeing which ones satisfy the equation. Since 11 is not too large, that
might be manageable. But before jumping into that, maybe there's a smarter way.

I remember something about quadratic equations modulo primes. There's this thing called the discriminant. For a
general quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\). If the discriminant is a
square modulo the prime, then there are solutions; otherwise, there aren't.

So in this case, let's compute the discriminant. Here, \(a = 1\), \(b = 1\), and \(c = 1\), so:

\(D = (1)^2 - 4(1)(1) = 1 - 4 = -3\).

Hmm, okay, so we have a discriminant of \(-3\) modulo 11. I need to check if \(-3\) is a quadratic residue modulo
11. In other words, does there exist some number \(y\) such that \(y^2 \equiv -3 \mod{11}\)?

Let me compute \(-3\) mod 11 first. Since 11 - 3 = 8, so \(-3 \equiv 8 \mod{11}\). So I need to check if 8 is a
quadratic residue modulo 11.

To do that, one method is to use Euler's criterion, which says that for an odd prime \(p\), and integer \(a\) not
divisible by \(p\), \(a^{(p-1)/2} \equiv 1 \mod{p}\) if \(a\) is a quadratic residue, and \(-1\) otherwise.

So let me compute \(8^{5} \mod{11}\). Why 5? Because (11 - 1)/2 = 5. Let's calculate:

\(8^1 = 8\)

\(8^2 = 64\), which is 64 - 5*11=64-55=9 mod 11.

\(8^3 = 8 * 9 = 72\). 72 divided by 11: 6*11=66, so 72-66=6. So \(8^3 \equiv 6 \mod{11}\).

\(8^4 = 8 * 6 = 48\), which is 48 - 4*11=48-44=4 mod 11.

\(8^5 = 8 * 4 = 32\). 32 - 2*11=32-22=10 mod 11. So \(8^5 \equiv 10 \mod{11}\).

Since 10 is not congruent to 1, Euler's criterion tells me that 8 is not a quadratic residue modulo 11. Therefore,
the discriminant isn't a square in this field, which suggests that there are no solutions.

Wait, but let me double-check because sometimes even if I think I'm correct, it's better to be thorough. Maybe
trying all possible residues from 0 to 10 will confirm whether any satisfy the equation \(x^2 + x + 1 \equiv 0
\mod{11}\).

Alright, let me create a table of values for each x in [0, 10] and compute \(x^2 + x + 1\) mod 11.

Starting with x=0:

\(0^2 + 0 + 1 = 1\). 1 mod 11 is 1. Not zero.

x=1:

\(1 + 1 + 1 = 3\). 3 mod 11 is 3. Nope.

x=2:

4 + 2 + 1 =7. 7 mod 11 is 7. Still not.

x=3:

9 + 3 + 1=13. 13-11=2. Not zero.

x=4:

16 + 4 + 1=21. 21 - 2*11=21-22=-1≡10 mod11. Still not zero.

x=5:

25 +5 +1=31. 31 -2*11=9. Not zero.

x=6:

36 +6 +1=43. 43 - 3*11=43-33=10. Not zero.

x=7:

49 +7 +1=57. 57 - 5*11=57-55=2. Nope.

x=8:

64 +8 +1=73. 73 -6*11=73-66=7. Not zero.

x=9:

81 +9 +1=91. 91 -8*11=91-88=3. Nope.

x=10:

100 +10 +1=111. 111 -10*11=111-110=1. Not zero.

Hmm, so none of the residues from 0 to 10 satisfy the equation \(x^2 + x + 1 \equiv 0 \mod{11}\). So that confirms
it: there are no solutions modulo 11 for this quadratic equation.

Wait a second, but I just remembered something. Quadratic equations over finite fields can have 0, 1, or 2
solutions. Since the discriminant wasn't a square, we have zero solutions here, which matches with our exhaustive
check above.

Alternatively, maybe there's another method involving completing the square? Let me try that for practice.

Starting with \(x^2 + x + 1 \equiv 0 \mod{11}\).

Completing the square:

\(x^2 + x = -1 \mod{11}\)

Take half of the coefficient of x, which is 1/2. Since we're working mod11, 1/2 is equivalent to 6 because
2*6=12≡1 mod11.

So, \( (x + 6)^2 - 36 ≡ -1 \mod{11}\).

Wait, let me compute that step by step:

\(x^2 + x = (x + 1/2)^2 - (1/2)^2\). So,

\( (x + 6)^2 - 36 ≡ -1 \mod{11} \)

Compute \( -36 mod11 \): 36 /11 is 3*11=33, so 36-33=3. Thus, -36≡-3≡8 mod11.

So:

\( (x + 6)^2 ≡ -1 + 36 mod11\), which simplifies to \( (x + 6)^2 ≡ (-1) + (36 mod11) \).

Wait, hold on. Let me redo that step because I might have messed up.

We had:

\( x^2 + x = -1 \mod{11} \)

Complete the square:

\( x^2 + x + (1/4) = -1 + (1/4) \mod{11} \)

So,

\( (x + 1/2)^2 ≡ (-1) + (1/4) mod11 \)

Compute \(1/4\) modulo11: since 4*3=12≡1 mod11, so 1/4≡3 mod11.

Thus:

Left side: \( (x + 6)^2 \mod{11} \) because 1/2 is 6 in mod11.

Right side: (-1) + 3 ≡ 2 mod11.

So the equation becomes:

\( (x + 6)^2 ≡ 2 mod11 \)

Now, so we need to solve \( y^2 ≡ 2 mod11 \), where y = x +6. So does 2 have square roots modulo11?

Let me check which squares give 2 mod11.

Compute squares of numbers from 0 to 10:

0²=0

1²=1

2²=4

3²=9

4²=16≡5

5²=25≡3

6²=36≡3

7²=49≡5

8²=64≡9

9²=81≡4

10²=100≡1.

So the squares modulo11 are 0,1,3,4,5,9. So 2 isn't a square modulo11 either. Therefore, equation \( y^2 ≡2
\mod{11}\) has no solution, so original equation also has no solutions.

Wait, but earlier when computing Euler's criterion for discriminant 8 (from the first method), we saw that 8 was
not a quadratic residue as well? Or wait, in the second method, we arrived at needing to solve y²≡2. So yeah, same
result: no solution.

So all methods converge on there being no solutions to this equation in integers mod11.

Therefore, I think it's settled. There are no integer solutions modulo11 for \( x^2 +x +1 ≡0 \mod{11} \).

**Final Answer**
There are no integer solutions. The final answer is \boxed{\text{No solution}}.
</think>

To determine the integer solutions of the quadratic equation \(x^2 + x + 1 \equiv 0 \pmod{11}\), we can use
several methods:

1. **Exhaustive Check**:
- We check all residues from 0 to 10 modulo 11.
- None of these values satisfy the equation \(x^2 + x + 1 \equiv 0 \pmod{11}\).

2. **Quadratic Residue Check**:
- The discriminant of the quadratic equation is calculated as \(\Delta = b^2 - 4ac = 1 - 4 = -3\).
- Converting \(-3\) modulo 11, we get \(8\). We check if 8 is a quadratic residue modulo 11.
- The squares modulo 11 are calculated: 0, 1, 3, 4, 5, 9. Since 8 is not among these, it is not a quadratic
residue.

3. **Completing the Square**:
- Rewrite the equation as \(x^2 + x = -1 \pmod{11}\).
- Complete the square: \((x + 6)^2 - 36 \equiv -1 \pmod{11}\).
- Simplifying, we get \((x + 6)^2 \equiv 2 \pmod{11}\). Checking if 2 is a quadratic residue modulo 11, we find
it is not.

Since all methods confirm that there are no solutions, we conclude:

\[
\boxed{\text{No solution}}
\]

我的本地版deepseek-r1:8b 为什么5秒就出结果了


>>> x^2 + x + 1 = 0 ( mod 11) 的整数解是啥
<think>
好，我现在遇到了一个同余方程的问题，需要解决它。题目是：x² + x + 1 ≡ 0 (mod 11)，我要找出满足这个方程的整数解。

首先，我回忆一下，模运算中的解法通常涉及到逐一测试可能的值，或者使用一些代数技巧来简化问题。这里，模数是11，所以x的
取值范围应该在0到10之间，因为模11下，余数只能是从0到10。

接下来，我想到可以将方程改写成一个更易处理的形式，可能会帮助我找到解。原方程是x² + x + 1 ≡ 0 (mod 11)，也就是x²
+ x ≡ -1 (mod 11)。因为-1在模11下等价于10，所以我可以写成x² + x ≡ 10 (mod 11)。

然后，我想到将这个方程转化为标准的二次方程形式，可能会有助于求解。标准形式是ax² + bx + c ≡ d (mod m)，这里a=1，
b=1，c=1，d=10，m=11。

接下来，我想到可以使用求根公式来解决模运算中的二次方程问题。在普通情况下，求根公式是x = [-b ± √(b² - 4ac)] /
(2a)。在这里，代入数值得到：

x = [-1 ± √(1 - 4*1*1)] / (2*1)
x = [-1 ± √(-3)] / 2

但是，这里出现了一个问题，因为-3在模11下不是一个完全平方数。我需要找到在模11下，哪些数的平方等于-3（即8）。于是，我
开始列出0到10之间所有数的平方，看看哪个等于8：

0² ≡ 0 (mod 11)
1² ≡ 1
2² ≡ 4
3² ≡ 9
4² ≡ 16 ≡ 5 (因为16 - 11 = 5)
5² ≡ 25 ≡ 3 (25 - 22 = 3)
6² ≡ 36 ≡ 3 (36 - 33 = 3)
7² ≡ 49 ≡ 5 (49 - 44 = 5)
8² ≡ 64 ≡ 9 (64 - 55 = 9)
9² ≡ 81 ≡ 4 (81 - 77 = 4)
10² ≡ 100 ≡ 1 (100 - 99 = 1)

看来在模11下，没有数的平方等于8。那么，原方程是否有解呢？这似乎没有解，但我不太确定。

为了验证这一点，我可以尝试逐一测试x从0到10，看看哪些值满足x² + x + 1 ≡ 0 (mod 11)。

让我列出所有可能的x值和对应的x² + x + 1：

- x = 0: 0 + 0 + 1 = 1 ≡ 1 ≠ 0
- x = 1: 1 + 1 + 1 = 3 ≡ 3 ≠ 0
- x = 2: 4 + 2 + 1 = 7 ≡ 7 ≠ 0
- x = 3: 9 + 3 + 1 = 13 ≡ 2 ≠ 0
- x = 4: 16 + 4 + 1 = 21 ≡ 10 ≠ 0
- x = 5: 25 + 5 + 1 = 31 ≡ 9 ≠ 0
- x = 6: 36 + 6 + 1 = 43 ≡ 10 ≠ 0
- x = 7: 49 + 7 + 1 = 57 ≡ 2 ≠ 0
- x = 8: 64 + 8 + 1 = 73 ≡ 73 - 66 = 7 ≠ 0
- x = 9: 81 + 9 + 1 = 91 ≡ 91 - 88 = 3 ≠ 0
- x = 10: 100 + 10 + 1 = 111 ≡ 111 - 110 = 1 ≠ 0

看来，所有x从0到10都没有满足方程的值，这意味着在模11下，这个方程没有解。

但是，我有点怀疑自己是否正确，因为有时候可能会漏掉某些情况或者计算错误。我再仔细检查一下每一个步骤。

首先，在转换方程时，确实得到了x² + x ≡ 10 (mod 11)。然后，使用求根公式得到√(-3)，但在模11下，-3等价于8，我需要找
到数y，使得y² ≡ 8 (mod 11)。

然而，从之前的计算可以看出，没有这样的y存在，因为0到10之间的平方数分别是0,1,4,9,5,3,3,5,9,4,1。因此，确实没有数满足
y² ≡ 8 (mod 11)，这意味着原方程在模11下没有解。

另外，我已经逐一测试了所有可能的x值，都没有满足x² + x + 1 ≡ 0 (mod 11)。因此，可以确认这个方程在模11下没有整数解。

</think>

根据上述思考过程，我们可以得出以下结论：

**步骤解析：**

1. **问题理解：**
我们需要解同余方程 \( x^2 + x + 1 \equiv 0 \pmod{11} \)，即找到满足该方程的整数 \( x \) 在模11下的值。

2. **代数变形：**
将原方程改写为：
\[
x^2 + x \equiv -1 \pmod{11}
\]
由于 \(-1 \equiv 10 \pmod{11}\)，方程变为：
\[
x^2 + x \equiv 10 \pmod{11}
\]

3. **求根公式应用：**
使用二次方程求解公式，得到：
\[
x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2} = \frac{-1 \pm \sqrt{-3}}{2}
\]
在模11下，\(-3 \equiv 8\)，所以需要找到满足 \( y^2 \equiv 8 \pmod{11} \) 的整数 \( y \)。

4. **平方剩余分析：**
计算模11下的平方值：
\[
0^2 = 0, \quad 1^2=1,\ 2^2=4,\ 3^2=9,\ 4^2=5,\ 5^2=3,\ 6^2=3,\ 7^2=5,\ 8^2=9,\ 9^2=4,\ 10^2=1
\]
发现没有数的平方等于8，即 \( 8 \) 不是模11下的平方剩余。

5. **验证解的存在性：**
因为没有满足 \( y^2 \equiv 8 \pmod{11} \) 的整数，因此原方程在模11下无解。

6. **逐一测试验证：**
逐个代入 \( x = 0,1,2,\ldots,10 \)，计算结果均不满足 \( x^2 + x + 1 \equiv 0 \pmod{11} \)。

**最终结论：**

在模11下，同余方程 \( x^2 + x + 1 \equiv 0 \pmod{11} \) 没有整数解。

[irisyuan]

我的本地版deepseek-r1:8b 为什么5秒就出结果了


>>> x^2 + x + 1 = 0 ( mod 11) 的整数解是啥
<think>
好，我现在遇到了一个同余方程的问题，需要解决它。题目是：x² + x + 1 ≡ 0 (mod 11)，我要找出满足这个方程的整数解。

首先，我回忆一下，模运算中的解法通常涉及到逐一测试可能的值，或者使用一些代数技巧来简化问题。这里，模数是11，所以x的
取值范围应该在0到10之间，因为模11下，余数只能是从0到10。

接下来，我想到可以将方程改写成一个更易处理的形式，可能会帮助我找到解。原方程是x² + x + 1 ≡ 0 (mod 11)，也就是x²
+ x ≡ -1 (mod 11)。因为-1在模11下等价于10，所以我可以写成x² + x ≡ 10 (mod 11)。

然后，我想到将这个方程转化为标准的二次方程形式，可能会有助于求解。标准形式是ax² + bx + c ≡ d (mod m)，这里a=1，
b=1，c=1，d=10，m=11。

接下来，我想到可以使用求根公式来解决模运算中的二次方程问题。在普通情况下，求根公式是x = [-b ± √(b² - 4ac)] /
(2a)。在这里，代入数值得到：

x = [-1 ± √(1 - 4*1*1)] / (2*1)
x = [-1 ± √(-3)] / 2

但是，这里出现了一个问题，因为-3在模11下不是一个完全平方数。我需要找到在模11下，哪些数的平方等于-3（即8）。于是，我
开始列出0到10之间所有数的平方，看看哪个等于8：

0² ≡ 0 (mod 11)
1² ≡ 1
2² ≡ 4
3² ≡ 9
4² ≡ 16 ≡ 5 (因为16 - 11 = 5)
5² ≡ 25 ≡ 3 (25 - 22 = 3)
6² ≡ 36 ≡ 3 (36 - 33 = 3)
7² ≡ 49 ≡ 5 (49 - 44 = 5)
8² ≡ 64 ≡ 9 (64 - 55 = 9)
9² ≡ 81 ≡ 4 (81 - 77 = 4)
10² ≡ 100 ≡ 1 (100 - 99 = 1)

看来在模11下，没有数的平方等于8。那么，原方程是否有解呢？这似乎没有解，但我不太确定。

为了验证这一点，我可以尝试逐一测试x从0到10，看看哪些值满足x² + x + 1 ≡ 0 (mod 11)。

让我列出所有可能的x值和对应的x² + x + 1：

- x = 0: 0 + 0 + 1 = 1 ≡ 1 ≠ 0
- x = 1: 1 + 1 + 1 = 3 ≡ 3 ≠ 0
- x = 2: 4 + 2 + 1 = 7 ≡ 7 ≠ 0
- x = 3: 9 + 3 + 1 = 13 ≡ 2 ≠ 0
- x = 4: 16 + 4 + 1 = 21 ≡ 10 ≠ 0
- x = 5: 25 + 5 + 1 = 31 ≡ 9 ≠ 0
- x = 6: 36 + 6 + 1 = 43 ≡ 10 ≠ 0
- x = 7: 49 + 7 + 1 = 57 ≡ 2 ≠ 0
- x = 8: 64 + 8 + 1 = 73 ≡ 73 - 66 = 7 ≠ 0
- x = 9: 81 + 9 + 1 = 91 ≡ 91 - 88 = 3 ≠ 0
- x = 10: 100 + 10 + 1 = 111 ≡ 111 - 110 = 1 ≠ 0

看来，所有x从0到10都没有满足方程的值，这意味着在模11下，这个方程没有解。

但是，我有点怀疑自己是否正确，因为有时候可能会漏掉某些情况或者计算错误。我再仔细检查一下每一个步骤。

首先，在转换方程时，确实得到了x² + x ≡ 10 (mod 11)。然后，使用求根公式得到√(-3)，但在模11下，-3等价于8，我需要找
到数y，使得y² ≡ 8 (mod 11)。

然而，从之前的计算可以看出，没有这样的y存在，因为0到10之间的平方数分别是0,1,4,9,5,3,3,5,9,4,1。因此，确实没有数满足
y² ≡ 8 (mod 11)，这意味着原方程在模11下没有解。

另外，我已经逐一测试了所有可能的x值，都没有满足x² + x + 1 ≡ 0 (mod 11)。因此，可以确认这个方程在模11下没有整数解。

</think>

根据上述思考过程，我们可以得出以下结论：

**步骤解析：**

1. **问题理解：**
我们需要解同余方程 \( x^2 + x + 1 \equiv 0 \pmod{11} \)，即找到满足该方程的整数 \( x \) 在模11下的值。

2. **代数变形：**
将原方程改写为：
\[
x^2 + x \equiv -1 \pmod{11}
\]
由于 \(-1 \equiv 10 \pmod{11}\)，方程变为：
\[
x^2 + x \equiv 10 \pmod{11}
\]

3. **求根公式应用：**
使用二次方程求解公式，得到：
\[
x = \frac{-1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2} = \frac{-1 \pm \sqrt{-3}}{2}
\]
在模11下，\(-3 \equiv 8\)，所以需要找到满足 \( y^2 \equiv 8 \pmod{11} \) 的整数 \( y \)。

4. **平方剩余分析：**
计算模11下的平方值：
\[
0^2 = 0, \quad 1^2=1,\ 2^2=4,\ 3^2=9,\ 4^2=5,\ 5^2=3,\ 6^2=3,\ 7^2=5,\ 8^2=9,\ 9^2=4,\ 10^2=1
\]
发现没有数的平方等于8，即 \( 8 \) 不是模11下的平方剩余。

5. **验证解的存在性：**
因为没有满足 \( y^2 \equiv 8 \pmod{11} \) 的整数，因此原方程在模11下无解。

6. **逐一测试验证：**
逐个代入 \( x = 0,1,2,\ldots,10 \)，计算结果均不满足 \( x^2 + x + 1 \equiv 0 \pmod{11} \)。

**最终结论：**

在模11下，同余方程 \( x^2 + x + 1 \equiv 0 \pmod{11} \) 没有整数解。

本地部署，怎么搞？需要什么level的硬件？这几天deepseek基本用不了，问就是server busy

[TheMatrix]

本地部署，怎么搞？需要什么level的硬件？这几天deepseek基本用不了，问就是server busy

viewtopic.php?t=699555

需要一个大内存的GPU。不过我没装，具体我也不知道。

[acongard]

我本地安装了最大的版本（70b），给出的答案是：

### Final Answer
The solutions are \(\boxed{4}\) and \(\boxed{8}\) modulo 11.

[TheMatrix]

本地部署，怎么搞？需要什么level的硬件？这几天deepseek基本用不了，问就是server busy