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====================

User prompt 1 of 8 - 10/9/2025, 1:08:11 AM:
hi

GPT-OSS 120B:
Hello! How can I help you today?

--------------------

User prompt 2 of 8 - 10/9/2025, 1:10:08 AM:
I need to check that e^k - t % p === 0. All numbers are known. e and p are primes. But k is large (e.g. 1024 bit). How can I perform the check without expanding the expression?

GPT-OSS 120B:
## Checking \(e^{k} \equiv t \pmod p\) without expanding \(e^{k}\)

When \(k\) is thousands of bits, computing the full integer \(e^{k}\) is infeasible.  
Instead use **modular exponentiation**, which repeatedly reduces intermediate results modulo \(p\).

### Algorithm (binary exponentiation)

1. **Input**: primes \(e, p\); integers \(k, t\) with \(0 \le t < p\).  
2. **Initialize**  
   * `result ← 1`  
   * `base ← e mod p`  
3. **Loop** while \(k > 0\)  
   * If the least‑significant bit of \(k\) is 1: `result ← (result * base) mod p`  
   * `base ← (base * base) mod p`   // square for the next bit  
   * Shift \(k\) right by one bit (`k ← k >> 1`).  
4. After the loop, `result` equals \(e^{k} \bmod p\).  
5. **Check**: `result == t`. If true, then \(e^{k} - t\) is divisible by \(p\).

### Why it works
At each step we keep the intermediate value reduced modulo \(p\).  
Because modular reduction is compatible with multiplication:

\[
(a \bmod p)\,(b \bmod p) \bmod p = (ab) \bmod p,
\]

the final `result` is exactly \(e^{k} \bmod p\) even though we never form the full power.

### Practical notes

| Consideration | Recommendation |
|---|---|
| **Large exponent** | Represent \(k\) as a binary string (e.g., a `bigint` in most languages). |
| **Performance** | The loop runs in \(O(\log k)\) multiplications; each multiplication is modulo‑\(p\) and can use fast algorithms (Montgomery reduction, Karatsuba) for very large \(p\). |
| **Constant‑time** | For cryptographic contexts, use a constant‑time implementation to avoid timing leaks. |
| **Libraries** | Most languages provide this directly: <br>• Python: `pow(e, k, p)` <br>• JavaScript (BigInt): `modPow(e, k, p)` via libraries like `bigint-mod-arith` <br>• C/C++: OpenSSL’s `BN_mod_exp`. |

### Example (Python)

```python
def check_power(e: int, k: int, t: int, p: int) -> bool:
    """Return True iff e**k ≡ t (mod p)."""
    return pow(e, k, p) == t

# sample values
e = 1019          # prime
p = 10007         # prime
k = int('1' * 1024, 2)   # 1024‑bit exponent (example)
t = 1234

print(check_power(e, k, t, p))
```

The built‑in `pow` performs modular exponentiation efficiently, handling the 1024‑bit exponent without ever constructing `e**k`.

--------------------

User prompt 3 of 8 - 10/9/2025, 1:21:57 AM:
I need to check that e^2k - t % p(e^k) === 0. All numbers are known. e and p are primes. But k is large (e.g. 1024 bit). How can I perform the check without expanding the expression?

GPT-OSS 120B:
## Goal  

Verify  

\[
e^{2k}-t \equiv 0 \pmod{p\,e^{k}}
\]

with  

* \(e,\;p\) – known primes  
* \(k\) – a 1024‑bit (or larger) integer  
* \(t\) – known integer  

without ever forming the gigantic number \(e^{2k}\).

---

## 1. Rewrite the condition  

Let  

\[
a \;=\; e^{k}.
\]

Then  

\[
e^{2k}=a^{2},
\qquad
p\,e^{k}=p\,a .
\]

The congruence becomes  

\[
a^{2} - t \equiv 0 \pmod{p\,a}
\;\Longleftrightarrow\;
a^{2} \equiv t \pmod{p\,a}.
\]

A modulus that is a product of two coprime factors (\(a\) and \(p\) are coprime because \(p\) is prime and \(p\neq e\)) lets us split the condition into two separate congruences:

| Modulus | Required congruence |
|---|---|
| \(a\) | \(a^{2} \equiv t \pmod a\) |
| \(p\) | \(a^{2} \equiv t \pmod p\) |

### 1.1. Modulo \(a\)

\(a^{2} \equiv t \pmod a\) reduces to  

\[
t \equiv 0 \pmod a \quad\Longrightarrow\quad a \mid t .
\]

So **\(a\) must divide \(t\)**. Write  

\[
t = a \cdot s
\]

for some integer \(s\).

### 1.2. Modulo \(p\)

Substituting \(t = a s\) into the second congruence:

\[
a^{2} \equiv a s \pmod p
\;\Longleftrightarrow\;
a \equiv s \pmod p .
\]

Thus **\(s \equiv a \pmod p\)**.

---

## 2. What we actually need to compute  

1. **\(a \bmod p\)** – the value of \(e^{k}\) modulo the (small) prime \(p\).  
2. **Divisibility test** – does \(a\) divide \(t\)? (i.e. is \(t \bmod a = 0\)?)  
3. If the division succeeds, compute \(s = t / a\) and check \(s \bmod p = a \bmod p\).

All of these steps avoid constructing \(e^{2k}\) or even \(e^{k}\) in full size; only the **mod‑\(p\)** value of \(e^{k}\) and a division by the (potentially huge) integer \(a\) are required.

---

## 3. Algorithm (binary modular exponentiation)

```python
def check_condition(e: int, k: int, t: int, p: int) -> bool:
    """
    Return True iff  e^(2k) - t is divisible by p * e^k.
    """
    # 1. Compute a_mod = e^k (mod p) using fast modular exponentiation
    a_mod = pow(e, k, p)          # O(log k) multiplications, all modulo p

    # 2. Compute a = e^k as a big integer only if needed for the division test.
    #    We can obtain it with the same exponentiation but without modulus.
    #    Many libraries provide a "bigint pow" that returns the full integer.
    a = pow(e, k)                 # may be huge (1024‑bit exponent → ~k*log2(e) bits)

    # 3. Check that a divides t
    if t % a != 0:
        return False

    # 4. Compute s = t // a and compare modulo p
    s = t // a
    return (s % p) == a_mod
```

### Remarks on step 2  

*If the language can handle the full integer `a = e**k` (e.g. Python’s arbitrary‑precision `int`), the division `t % a` is straightforward.*  
*If the full integer is still too large for the environment, you can perform the divisibility test using **modular reduction with the same modulus** `a` via the Euclidean algorithm on the binary representation of `k`. Many big‑integer libraries expose a “mod‑by‑big‑int” routine that works without materialising the entire power.*

---

## 4. Complexity  

| Operation | Cost |
|---|---|
| Compute \(e^{k} \bmod p\) | \(O(\log k)\) modular multiplications (each \(O(\log^{2} p)\) with schoolbook multiplication) |
| Compute full \(e^{k}\) (if needed) | \(O(\log k)\) multiplications on numbers that grow to \(\approx k\log_{2}e\) bits |
| Division `t % a` | \(O(\text{size}(t) \times \text{size}(a))\) using standard big‑int division; still feasible for a few thousand‑bit numbers |
| Final modulo‑\(p\) check | negligible |

The dominant work is the modular exponentiation, which is fast even for a 1024‑bit exponent.

---

## 5. Summary of the check  

1. Compute \(a_{\!p}=e^{k}\bmod p\).  
2. Compute (or obtain) the full integer \(a=e^{k}\).  
3. Verify that \(a\) divides \(t\). If not, the condition fails.  
4. Let \(s = t / a\).  
5. Accept iff \(s \bmod p = a_{\!p}\).

All steps avoid forming \(e^{2k}\) and rely only on modular exponentiation and a single division, making the verification practical even when \(k\) is thousands of bits long.

--------------------

User prompt 4 of 8 - 10/9/2025, 1:26:57 AM:
I would like to test if there are solutions to e^k % (px + t) === 0. e and p are primes, k could be large.

GPT-OSS 120B:
## When can \(e^{k}\) be divisible by a linear expression?

You want to know whether there exists an integer \(k\ge 1\) such that  

\[
e^{k}\equiv 0 \pmod{\,p\,x+t\,},
\]

where  

* \(e\) – prime (the base of the exponent)  
* \(p\) – prime (coefficient of \(x\))  
* \(t\) – known integer (constant term)  
* \(x\) – an integer variable (you may treat it as a free parameter)  

The modulus is the linear polynomial \(M = p\,x + t\).  
For a congruence of the form \(e^{k}\equiv 0 \pmod M\) to hold, **\(M\) must divide the integer \(e^{k}\)**.

### 1. Divisibility condition

If a positive integer \(d\) divides \(e^{k}\), then every prime factor of \(d\) must also be a factor of \(e\).  
Since \(e\) itself is prime, the only prime divisor of any power \(e^{k}\) is \(e\) itself. Consequently

\[
d \mid e^{k}\quad\Longrightarrow\quad d = e^{m}\quad\text{for some }0\le m\le k .
\]

Thus **any divisor of \(e^{k}\) is a power of \(e\)** (including \(1 = e^{0}\)).

### 2. Apply to the linear modulus

The modulus \(M = p\,x + t\) can be any integer depending on the choice of \(x\).  
For the congruence to be solvable we need

\[
p\,x + t = e^{m}\qquad\text{for some }m\ge 0 .
\]

That is, the linear expression must itself be an exact power of the prime \(e\).

### 3. Existence test

1. **Check whether \(t\) is already a power of \(e\).**  
   If \(t = e^{m}\) for some \(m\), then setting \(x = 0\) gives \(M = t = e^{m}\) and the congruence holds with any \(k\ge m\).

2. **Otherwise, solve for \(x\).**  
   Rearrange  

   \[
   p\,x = e^{m} - t \;\Longrightarrow\; x = \frac{e^{m} - t}{p}.
   \]

   For a given exponent \(m\) you obtain an integer \(x\) **iff** the right‑hand side is divisible by \(p\).  
   So the problem reduces to finding an exponent \(m\) such that  

   \[
   e^{m} \equiv t \pmod p .
   \]

   This is a discrete‑logarithm problem: find \(m\) with  

   \[
   e^{m} \equiv t \pmod p .
   \]

   If a solution \(m\) exists, compute  

   \[
   x = \frac{e^{m} - t}{p},
   \]

   and then any \(k\ge m\) satisfies the original condition because \(e^{k}\) is a multiple of \(e^{m}=p\,x+t\).

### 4. Practical steps

| Step | Action |
|---|---|
| **A** | Compute \(t \bmod p\). If \(t \equiv 0\) then \(M = p\,x\) can never be a power of \(e\) (since \(e\) is prime \(\neq p\)). No solution. |
| **B** | Solve the discrete log \(e^{m} \equiv t \pmod p\). Use algorithms such as Baby‑Step‑Giant‑Step (complexity \(O(\sqrt{p})\)) or Pollard’s rho for logarithms. |
| **C** | If no \(m\) is found, there is **no** integer \(x\) making \(p\,x+t\) a power of \(e\); the original congruence has no solution. |
| **D** | If a solution \(m\) is found, compute \(x = (e^{m} - t)/p\). Verify that \(x\) is an integer (the division will be exact by construction). |
| **E** | Choose any \(k \ge m\); then \(e^{k}\) is divisible by \(p\,x+t\). |

### 5. Example

Suppose  

* \(e = 5\) (prime)  
* \(p = 7\) (prime)  
* \(t = 3\).

1. Solve \(5^{m} \equiv 3 \pmod 7\).  
   * \(5^{1}=5 \equiv 5\)  
   * \(5^{2}=25 \equiv 4\)  
   * \(5^{3}=125 \equiv 6\)  
   * \(5^{4}=625 \equiv 2\)  
   * \(5^{5}=3125 \equiv 3\) → \(m=5\).

2. Compute \(x = (5^{5} - 3)/7 = (3125-3)/7 = 3122/7 = 446\).

3. Then \(p\,x + t = 7·446 + 3 = 3125 = 5^{5}\).  

4. Any \(k \ge 5\) works; e.g. \(k=5\) gives \(5^{5}=3125\) which is exactly divisible by \(3125\).

### 6. Summary

* The congruence \(e^{k} \equiv 0 \pmod{p\,x+t}\) is solvable **iff** the linear term \(p\,x+t\) equals a power of the base prime \(e\).  
* This reduces to a discrete‑logarithm problem modulo \(p\): find \(m\) with \(e^{m}\equiv t\pmod p\).  
* If such an \(m\) exists, set \(x = (e^{m}-t)/p\) and any \(k\ge m\) satisfies the original condition.  
* If no discrete‑log solution exists, there are no integer solutions for \(x\) and \(k\).

--------------------

User prompt 5 of 8 - 10/9/2025, 1:31:03 AM:
I would like to test if there are solutions to e^2k % (e^k*p*x + t) === 0. e and p are primes, k could be large.

GPT-OSS 120B:
## Problem restated  

Find whether there exist integers  

* \(k\ge 1\) (possibly thousands of bits)  
* \(x\) (any integer)  

such that  

\[
e^{2k}\equiv 0 \pmod{\,e^{k}\,p\,x + t\,},
\]

where  

* \(e\) – prime (base of the exponent)  
* \(p\) – prime (coefficient of \(x\))  
* \(t\) – known integer  

The modulus is  

\[
M = e^{k}\,p\,x + t .
\]

The congruence means **\(M\) must divide \(e^{2k}\)**.

---

## 1. Divisibility condition

Since \(e^{2k}= (e^{k})^{2}\) is a pure power of the prime \(e\), every divisor of \(e^{2k}\) is itself a power of \(e\):

\[
d\mid e^{2k}\quad\Longrightarrow\quad d = e^{m}\quad\text{for some }0\le m\le 2k .
\]

Therefore we require  

\[
e^{k}\,p\,x + t = e^{m} \qquad\text{for some }m\le 2k .
\tag{1}
\]

---

## 2. Solving equation (1)

Rearrange (1) to isolate the term containing \(x\):

\[
p\,x = e^{m} - t\;e^{-k}.
\]

Because \(e^{k}\) is a factor of the left‑hand side, the right‑hand side must be an **integer multiple of \(e^{k}\)**. Write  

\[
e^{m} = e^{k}\,q \qquad\text{with } q = e^{m-k}\in\mathbb{Z}.
\]

Substituting:

\[
p\,x = e^{k}\,q - t .
\tag{2}
\]

Equation (2) shows that we need an integer \(q\) (a power of \(e\)) such that  

\[
e^{k}\,q \equiv t \pmod p .
\tag{3}
\]

If (3) holds, then  

\[
x = \frac{e^{k}\,q - t}{p}
\]

is an integer and (1) is satisfied with \(m = k + \log_{e} q\).

---

## 3. Reducing to a discrete‑logarithm

Let  

\[
q = e^{r}\quad (r\ge 0).
\]

Then (3) becomes  

\[
e^{k+r} \equiv t \pmod p .
\]

Define  

\[
s = k+r .
\]

We need a **single exponent \(s\)** (with \(s\ge k\)) such that  

\[
e^{s} \equiv t \pmod p .
\tag{4}
\]

Finding such an \(s\) is exactly the **discrete logarithm problem** modulo the prime \(p\).

*If a solution \(s\) exists*, set  

\[
r = s-k\ (\ge 0),\qquad q = e^{r},\qquad
x = \frac{e^{k}\,e^{r} - t}{p}
      = \frac{e^{s} - t}{p}.
\]

Because \(e^{s}\equiv t\pmod p\), the numerator is a multiple of \(p\), so \(x\) is integral.  
Finally, \(m = k+r = s\) and \(M = e^{m}\) divides \(e^{2k}\) (since \(m\le 2k\) whenever \(s\le 2k\); if the found \(s\) exceeds \(2k\) you can increase \(k\) accordingly).

---

## 4. Algorithmic steps

| Step | Action |
|---|---|
| **A** | Compute \(t \bmod p\). |
| **B** | Solve the discrete log \(e^{s} \equiv t \pmod p\). Use Baby‑Step‑Giant‑Step (cost \(O(\sqrt{p})\)) or Pollard‑Rho for logarithms. |
| **C** | If no \(s\) exists → **no solution** for any \(k,x\). |
| **D** | Choose any \(k\) such that \(k \le s \le 2k\). The simplest is to set \(k = \lceil s/2\rceil\). |
| **E** | Compute \(x = (e^{s} - t)/p\) (the division is exact by construction). |
| **F** | Verify: \(M = e^{k}p x + t = e^{s}\) and indeed \(e^{2k}\) is a multiple of \(e^{s}\). |

### Why the bound \(s\le 2k\) matters
We need \(M\) to be a divisor of \(e^{2k}\). Since \(M = e^{s}\), this requires \(s \le 2k\). By picking \(k\) as at least \(\lceil s/2\rceil\) we guarantee the condition.

---

## 5. Example (small numbers)

Take  

* \(e = 5\) (prime)  
* \(p = 7\) (prime)  
* \(t = 3\).

1. Solve \(5^{s} \equiv 3 \pmod 7\).  
   Computing powers: \(5^{5}=3125\equiv 3\pmod7\). So \(s=5\).  
2. Choose \(k = \lceil 5/2\rceil = 3\). Then \(2k = 6 \ge s\).  
3. Compute \(x = (5^{5} - 3)/7 = (3125-3)/7 = 3122/7 = 446\).  
4. Check \(M = 5^{3}\cdot7\cdot446 + 3 = 125\cdot3122 + 3 = 390250 + 3 = 390253 = 5^{5}\).  
5. \(e^{2k}=5^{6}=15625\) is divisible by \(5^{5}=390253\)?  
   Oops – my arithmetic shows \(5^{5}=3125\), not 390253.  
   Correct calculation:  

   *\(e^{k}=5^{3}=125\)*  
   *\(M = 125\cdot7\cdot446 + 3 = 125\cdot3122 + 3 = 390250 + 3 = 390253\).*  

   This is **not** a power of 5, so the example shows that the discrete‑log solution must also satisfy the exact equality \(M = e^{s}\).  

   The mistake arises because we used the same \(s\) from the congruence but did not enforce that \(M\) equals \(e^{s}\) exactly; the extra factor \(e^{k}\) in \(M\) changes the value.  

   The correct approach is to set \(k\) **equal to** the exponent found: take \(k = s = 5\). Then  

   *\(e^{k}=5^{5}=3125\)*  
   *\(x = (5^{5} - 3)/(5^{5}p) = (3125-3)/(3125\cdot7) = 3122/21875\) – not integral, so no integer \(x\).  

   This illustrates that a solution exists **only when** the discrete‑log exponent \(s\) also makes \((e^{s}-t)\) divisible by \(p\) **and** the quotient \((e^{s}-t)/p\) is a multiple of \(e^{k}\).  

   In practice you would iterate over possible \(k\) (or directly over \(s\)) until both divisibility conditions hold.

---

## 6. Practical procedure (search over \(k\))

Because the extra factor \(e^{k}\) appears in the modulus, a robust method is:

1. **Find all solutions** \(s\) of the discrete log \(e^{s}\equiv t\pmod p\).  
2. For each such \(s\), try to write  

   \[
   e^{s} - t = p\,e^{k}\,x
   \]

   i.e. check whether \((e^{s}-t)/p\) is **exactly divisible by a power of \(e\)**.  
   Compute  

   \[
   v = \frac{e^{s}-t}{p}.
   \]

   Factor out powers of \(e\) from \(v\): let \(v = e^{k}\,x\) with \(x\) not divisible by \(e\).  
   If such a decomposition exists (i.e. \(v\) is not zero), then the pair \((k,x)\) satisfies the original condition.

3. Verify that \(k\le s\le 2k\) (

--------------------

User prompt 6 of 8 - 10/9/2025, 1:47:42 AM:
I would like to test if there are solutions to e % (px + t) === 0. e is a power of a prime, p is a prime. e could be large

GPT-OSS 120B:
## When can a prime‑power be divisible by a linear expression?

You want to know whether there exist integers  

* \(x\) (any integer)  

such that  

\[
e \equiv 0 \pmod{p\,x + t},
\]

where  

* \(e = q^{\,n}\) is a power of a prime \(q\) (so \(q\) may equal \(p\) or be different)  
* \(p\) is a prime (coefficient of \(x\))  
* \(t\) is a known integer  

The congruence means **\(p\,x + t\) must divide \(e\)**.

---

### 1. Divisors of a prime‑power

All positive divisors of \(e = q^{\,n}\) are themselves powers of the same prime \(q\):

\[
d \mid q^{\,n}\quad\Longrightarrow\quad d = q^{\,m}\ \text{for some }0\le m\le n .
\]

Therefore the linear expression can be a divisor **iff it is exactly a power of \(q\)**.

---

### 2. Required form of the linear term

We need  

\[
p\,x + t = q^{\,m}\qquad (0\le m\le n).
\tag{1}
\]

Solve (1) for \(x\):

\[
x = \frac{q^{\,m} - t}{p}.
\tag{2}
\]

Thus a solution exists **iff** there is an exponent \(m\) such that the numerator in (2) is an integer multiple of \(p\).

---

### 3. How to test it

1. **Iterate over possible exponents** \(m = 0,1,\dots ,n\).  
   For each \(m\) compute \(q^{\,m}\) (modular exponentiation is cheap even for large \(n\) because the numbers stay relatively small—\(q^{\,m}\) grows at most to the size of \(e\)).  

2. Check the divisibility condition  

   \[
   (q^{\,m} - t) \bmod p = 0 .
   \]

   If it holds, set  

   \[
   x = (q^{\,m} - t)/p
   \]

   and you have a valid solution.

3. If no exponent \(m\) satisfies the condition, **no integer \(x\) exists**.

Because \(m\) ranges only up to \(n\) (the exponent of the known prime‑power), the search is bounded by the size of the input, not by the magnitude of \(e\) itself.

---

### 4. Example

* \(e = 2^{10} = 1024\) (so \(q=2,\ n=10\))  
* \(p = 7\) (prime)  
* \(t = 3\)

Check exponents:

| \(m\) | \(q^{m}\) | \(q^{m}-t\) | \((q^{m}-t)\bmod 7\) | Result |
|---|---|---|---|---|
| 0 | 1 | -2 | 5 | ✗ |
| 1 | 2 | -1 | 6 | ✗ |
| 2 | 4 | 1  | 1 | ✗ |
| 3 | 8 | 5  | 5 | ✗ |
| 4 | 16| 13 | 6 | ✗ |
| 5 | 32| 29 | 1 | ✗ |
| 6 | 64| 61 | 5 | ✗ |
| 7 |128|125 | 6 | ✗ |
| 8 |256|253 | 1 | ✗ |
| 9 |512|509 | 5 | ✗ |
|10 |1024|1021| 1 | ✗ |

No exponent yields a multiple of 7, so no integer \(x\) satisfies the original congruence.

---

### 5. Summary

* A divisor of a prime‑power \(e = q^{\,n}\) must itself be a power of the same prime \(q\).  
* Therefore \(p\,x + t\) can divide \(e\) **iff** it equals \(q^{\,m}\) for some \(0\le m\le n\).  
* Test each exponent \(m\) by checking whether \(q^{\,m} \equiv t \pmod p\).  
* If such an \(m\) exists, compute \(x = (q^{\,m} - t)/p\); otherwise, no solution exists.

--------------------

User prompt 7 of 8 - 10/9/2025, 1:50:33 AM:
can i avoid iteration?

GPT-OSS 120B:
### Avoiding a naïve scan of all exponents  

The condition  

\[
p\,x + t = q^{\,m}\qquad(0\le m\le n)
\]

is equivalent to  

\[
q^{\,m} \equiv t \pmod p .
\tag{★}
\]

So the problem reduces to **finding an exponent \(m\) that satisfies the congruence (★)**.  
That is exactly a **discrete‑logarithm problem** modulo the prime \(p\).

If you can solve the discrete log, you obtain the smallest (or any) solution \(m\) without iterating through every exponent.

---

## 1. Discrete‑log solution  

Define the base \(g = q \bmod p\) (if \(q\equiv 0\pmod p\) then the only possible divisor is \(p\) itself, which is handled separately).  
We need an integer \(m\) such that  

\[
g^{\,m} \equiv t \pmod p .
\]

Standard algorithms:

| Algorithm | Complexity | When to use |
|---|---|---|
| **Baby‑Step‑Giant‑Step** | \(O(\sqrt{p})\) time, \(O(\sqrt{p})\) memory | Small‑to‑moderate primes (up to ≈ 2⁴⁰) |
| **Pollard’s ρ for logarithms** | \(O(\sqrt{p})\) expected time, negligible memory | Large primes where memory is a concern |
| **Pohlig‑Hellman** (if \(p-1\) is smooth) | \(O(\sum \log p_i)\) where \(p_i\) are prime factors of \(p-1\) | When \(p-1\) has small factors |
| **Index‑calculus / Number‑field sieve** | Sub‑exponential, used for very large \(p\) (≥ 2⁶⁰) | Specialized libraries (e.g., CADO‑NFS) |

All of these return the **discrete logarithm** \(m\) (or report that none exists).  

---

## 2. From \(m\) to \(x\)

Once you have a solution \(m\):

1. Compute the exact power \(q^{\,m}\) (big‑integer arithmetic).  
2. Verify that \(q^{\,m} \ge t\) and that \((q^{\,m}-t)\) is divisible by \(p\).  
3. Set  

   \[
   x = \frac{q^{\,m} - t}{p}.
   \]

If the division is integral, you have a valid pair \((x,m)\).  
If the discrete‑log algorithm returns a solution but the division fails, it means the congruence held modulo \(p\) but the exact equality \(p\,x+t = q^{\,m}\) cannot be satisfied; in that case **no solution exists**.

---

## 3. Edge cases  

* **\(q \equiv 0 \pmod p\)** – then \(p\) divides every power of \(q\).  
  *If \(t\equiv 0\pmod p\) the expression \(p\,x+t\) is a multiple of \(p\) and can equal \(q^{\,m}\) only when \(t=0\). Otherwise no solution.*

* **\(t\) not coprime to \(p\)** – the discrete‑log equation may have no solution because the right‑hand side must lie in the multiplicative subgroup generated by \(q\) modulo \(p\). The algorithms will simply report “no logarithm”.

---

## 4. Summary workflow (no explicit iteration)

1. Reduce the base: \(g = q \bmod p\).  
2. Solve the discrete logarithm \(g^{\,m} \equiv t \pmod p\) with an appropriate algorithm.  
3. If a solution \(m\) is found, compute \(q^{\,m}\) and set \(x = (q^{\,m}-t)/p\).  
4. If the division is integral, the original congruence holds; otherwise, there is no solution.

Thus the search over all exponents is replaced by a single discrete‑log computation, which is the standard, non‑iterative method for this class of problems.

--------------------

User prompt 8 of 8 - 10/9/2025, 1:51:14 AM:
can i avoid discrete log, since i don't need to know m

GPT-OSS 120B:
### Goal  
Find out whether **any** exponent \(m\) (with \(0\le m\le n\)) exists such that  

\[
q^{\,m}\equiv t \pmod p ,
\]

where  

* \(e = q^{\,n}\) is a known prime‑power,  
* \(p\) is a prime,  
* \(t\) is a known integer.  

You do **not** need the actual value of \(m\); you only need a yes/no answer.

---

## 1. The underlying group‑theoretic condition  

Modulo a prime \(p\) the non‑zero residues form the cyclic multiplicative group  

\[
\mathbb{F}_p^{\times}\; \cong\; C_{p-1}.
\]

Let  

\[
g = q \bmod p .
\]

If \(g\equiv 0\pmod p\) the only possible divisor of \(e\) that is a multiple of \(p\) is \(p\) itself; in that case the congruence can hold **only** when \(t\equiv 0\pmod p\).  
Assume from now on \(g\neq 0\).

The set of powers of \(g\) is the **subgroup**  

\[
\langle g\rangle = \{g^{0},g^{1},\dots ,g^{\operatorname{ord}(g)-1}\},
\]

where \(\operatorname{ord}(g)\) is the **multiplicative order** of \(g\) modulo \(p\) (the smallest positive \(r\) with \(g^{r}\equiv1\pmod p\)).

A solution \(m\) exists **iff** the target residue \(t\bmod p\) lies in that subgroup:

\[
t \bmod p \in \langle g\rangle .
\tag{★}
\]

Thus we only have to test membership, not to compute the exponent.

---

## 2. Membership test without discrete logarithm  

A residue belongs to \(\langle g\rangle\) exactly when raising it to the **order of the subgroup** yields 1:

\[
t^{\operatorname{ord}(g)} \equiv 1 \pmod p .
\tag{1}
\]

*Proof sketch*:  
If \(t = g^{m}\) for some \(m\), then \(t^{\operatorname{ord}(g)} = (g^{m})^{\operatorname{ord}(g)} = g^{m\operatorname{ord}(g)} \equiv (g^{\operatorname{ord}(g)})^{m} \equiv 1^{m}=1\).  
Conversely, in a cyclic group the set of elements whose \(\operatorname{ord}(g)\)-th power is 1 is exactly the subgroup generated by \(g\).

Hence the existence question reduces to:

1. **Compute \(\operatorname{ord}(g)\).**  
2. **Check** whether \(t^{\operatorname{ord}(g)} \equiv 1 \pmod p\).

If the test succeeds, at least one exponent \(m\) exists (it will automatically satisfy \(m\le n\) as long as \(n\ge \operatorname{ord}(g)\); if \(n\) is smaller you can additionally verify that the smallest such \(m\) does not exceed \(n\) – see the note below).

---

## 3. Computing the order efficiently  

The order of an element in \(\mathbb{F}_p^{\times}\) divides \(p-1\).  
Factor \(p-1\) (or at least obtain its prime‑power decomposition).  
If  

\[
p-1 = \prod_{i=1}^{r} \ell_i^{\,e_i},
\]

then the order can be found by the standard “divide‑out” method:

```python
def order_mod(g, p, factorization):
    # factorization = [(ℓ1, e1), (ℓ2, e2), ...] of p-1
    ord = p-1
    for ℓ, e in factorization:
        for _ in range(e):
            if pow(g, ord // ℓ, p) == 1:
                ord //= ℓ
    return ord
```

*Complexity*: \(O(\log p)\) modular exponentiations, each \(O(\log^2 p)\) time.  
If full factorization of \(p-1\) is too costly, you can use a **partial factorization**: the algorithm still yields the exact order as long as every prime factor of the true order is represented in the list; missing factors will simply leave the order larger, which makes the membership test (1) **conservative** (it may give a false‑negative but never a false‑positive).  

---

## 4. Handling the bound \(m\le n\)

The test above guarantees the *existence* of some exponent \(m\).  
If you also need to ensure that the exponent does not exceed the known exponent \(n\) of the prime‑power \(e = q^{\,n}\), proceed as follows:

1. After confirming membership, compute the **actual order** \(r = \operatorname{ord}(g)\).  
2. The set of possible exponents is the arithmetic progression  

   \[
   m \equiv m_{0} \pmod r,
   \]

   where \(m_{0}\) is the (unknown) discrete log.  
3. The smallest non‑negative solution is at most \(r-1\).  
4. If \(r-1 \le n\) then certainly a solution with \(m\le n\) exists.  
5. If \(r-1 > n\), you can still have a solution because the actual \(m_{0}\) might be smaller than \(r\). In that case you would need the discrete log to know the exact value; otherwise you can only state “a solution may exist, but we cannot guarantee it fits within the given exponent bound without further computation.”

So **without computing the discrete log you can definitively answer “yes” only when \(r-1 \le n\)**. Otherwise the answer is “unknown – a discrete‑log computation would be required to decide”.

---

## 5. Full decision procedure (no discrete log)

```text
Input: q, n, p (prime), t
1. g ← q mod p
2. if g = 0:
       return (t mod p == 0)   // only possible when t ≡ 0 (mod p)
3. Factor p‑1 (or obtain a sufficient partial factorization)
4. r ← order_mod(g, p, factorization)   // multiplicative order of g
5. if pow(t, r, p) != 1:
       return "No solution"
6. // at this point t lies in ⟨g⟩
   if r-1 ≤ n:
       return "Solution exists (guaranteed within exponent bound)"
   else:
       return "Solution may exist, but need discrete log to check bound"
```

All operations are modular exponentiations and integer arithmetic; no iteration over the exponent range and no explicit discrete‑log computation is performed.

---

### Bottom line
*Yes, you can avoid a full discrete‑log search.*  
Compute the order of \(q\) modulo \(p\) and test whether \(t^{\text{order}} \equiv 1 \pmod p\).  
If the order is small enough relative to the known exponent \(n\), you obtain a definitive “solution exists” answer; otherwise you can only state that a solution is possible but cannot confirm it without the actual logarithm.