<think>
Okay, so I'm trying to solve this problem where I need to find all primes \( p > 5 \) such that there exists an integer \( a \) and an integer \( r \) with \( 1 \leq r \leq p - 1 \). The key condition is that the sequence \( 1, a, a^2, \dots, a^{p-5} \) can be rearranged into another sequence \( b_0, b_1, b_2, \dots, b_{p-5} \) where for each \( n \) from 1 to \( p - 5 \), the difference \( b_n - b_{n-1} - r \) is divisible by \( p \).

Hmm, let's break this down. So essentially, after rearrangement, each consecutive pair in the new sequence has a constant difference modulo \( p \). That sounds like an arithmetic progression modulo \( p \), but not necessarily starting at 1.

Wait, so if I rearrange the powers of \( a \) into such a sequence, it must form an arithmetic progression with common difference \( r \) modulo \( p \).

But hold on, the original sequence is geometric: each term is multiplied by \( a \) to get the next one. After rearrangement, we're looking at it forming an arithmetic progression. So how can a geometric sequence also be rearranged into an arithmetic sequence?

That seems restrictive. Maybe only for certain primes or specific values of \( a \) and \( r \).

First, let's consider what is being asked: The powers of \( a \) modulo \( p \), when rearranged, form an arithmetic progression with difference \( r \). So, the set \( S = \{1, a, a^2, \dots, a^{p-5}\} \) must be such that there exists some permutation of these elements which is an arithmetic sequence modulo \( p \).

But how long is this sequence? The length is \( p - 4 \), since it goes from exponent 0 to \( p - 5 \). So the number of terms is \( (p - 5) + 1 = p - 4 \).

Wait, primes \( p > 5 \), so starting from \( p=7 \), sequences of length 3 up.

Let me try small primes first to see if I can spot a pattern or maybe get some intuition.

Take \( p = 7 \). Then the sequence is \(1, a, a^2\), which has length 3. We need to arrange them as an arithmetic progression with difference \( r \) modulo 7.

So possibilities for \( S = \{1, a, a^2\} \).

An arithmetic progression of length 3 can be either increasing or decreasing, and the middle term is the average of the first and third terms. So, if we have three elements in an arithmetic sequence, the middle one must equal the average of the other two modulo 7.

So, for example, let's choose \( a = 3 \). Then, \( S = \{1, 3, 2\} \) because \( 3^2 = 9 ≡ 2 mod 7 \).

So our set is {1, 2, 3}. Can this be arranged into an arithmetic progression? Yes, it can: 1, 2, 3. Here, each term differs by 1.

So here we have \( r = 1 \). So for \( p = 7 \), such an arrangement is possible with \( a=3 \) and \( r=1 \).

Wait, but let's check if that works. The original sequence: [1, 3, 2]. After rearrangement: [1, 2, 3]. Then, the differences are (2 - 1) = 1, which is \( r \mod 7 \), and (3 - 2) = 1, same as before. So that's correct.

Alternatively, maybe there are multiple \( a \) or \( r \). Let me try another \( a \).

Take \( a = 2 \). Then S = {1, 2, 4} because \( 2^2=4 mod 7 \). Now can these be arranged into an arithmetic progression? The possible orderings are:

- 1,2,3: but we don't have 3.

- 1,2,4: The differences would be 1 and 2. Not constant.

- 1,4, something: But with only three elements.

Wait, the set is {1, 2, 4}. Let's see if these can form an arithmetic progression.

The possible sequences:

Start at 1, difference r=1: 1, 2, 3. Doesn't have 3.

Difference r=2: 1, 3, 5. Doesn't match our set.

Wait, maybe starting from another element?

If we rearrange as 2,4,1, what's the differences:

4 - 2 = 2

1 - 4 = -3 ≡ 4 mod7. Not equal to 2.

Alternatively, 4,1,2: Difference is (-3) and 1, which is not same modulo 7.

Or 1,4,2: Differences are 3 and -2, which isn't consistent.

So in this case, with \( a=2 \), we can't rearrange to get an arithmetic sequence.

But when \( a=3 \), we could. So for p=7, there exists at least one such \( a \).

Hence, p=7 is acceptable.

Now let's check next prime: p=11.

So p=11, the sequence has exponents from 0 to 6, so terms are \(1, a, a^2, ..., a^6\). So S is of size 7 elements. We need to arrange these 7 elements into an arithmetic progression modulo 11 with common difference r.

An arithmetic progression of length 7 requires that the set S must be equally spaced mod 11.

Given that, the multiplicative order of \( a \) modulo 11 needs to be such that its powers cover the required residues. Hmm, but arranging them into an AP is tricky because in an AP, each term is linearly increasing, whereas the geometric sequence grows exponentially.

Wait, perhaps only certain primes p will satisfy this condition where the multiplicative group structure aligns with additive structures.

Alternatively, maybe for p=7 and 13?

But let me check p=11. Maybe try a primitive root mod 11. The primitive roots modulo 11 are 2,6,7,8.

Let's take \( a = 2 \). Then, the powers of 2 modulo 11 cycle as follows:

\( 2^0 =1\), \(2^1=2\), \(2^2=4\), \(2^3=8\), \(2^4=16≡5\), \(2^5=10\), \(2^6=20≡9\). So S={1,2,4,5,8,9,10}.

Wait, so that's 7 elements. Now can we arrange these into an arithmetic progression mod 11?

An AP of length 7 mod11 will have terms like c, c+r, c+2r,...c+6r mod11. So the set is {c + kr | k=0,…,6}.

So our S must be equal to that set for some c and r.

Alternatively, the elements of S should form a complete residue system modulo 7 with difference r?

Wait, but mod 11, so need to see if the multiplicative subgroup generated by 2 (in this case) is an additive shift of multiples of r.

But the set {1,2,4,5,8,9,10} does not look like a linear sequence. Let's check:

Let me list all elements: 1,2,4,5,8,9,10.

Is this an AP? Let's see if any constant difference can produce these numbers.

Looking at the differences between consecutive terms in S as sorted: 1,2,4,5,8,9,10

Differences are 1 (from 1-2), then 2 (2-4), then 1 (4-5), 3 (5-8), 1 (8-9), 1 (9-10). No, not consistent. So it's not an AP.

Alternatively, is there a permutation where the differences are constant?

Wait, let me see:

Suppose we try starting at 1 and adding some r each time. Let me choose r= something.

Looking at S={1,2,4,5,8,9,10}.

Trying to make an AP with these elements.

Let's compute pairwise differences:

From 1: possible next term is 2 (difference 1), or 4 (diff3), or 5(diff4), etc.

If I choose r=1:

AP would be 1,2,3,4,... but we don't have 3,6,7. So nope.

r=2: Starting at 1: 1,3,5,7,9,11≡0, which isn't in our set.

Not helpful.

Wait, another approach is to compute all possible consecutive differences within S and see if any r appears enough times.

But since we have a set of size 7, the number of required r's is 6. Each difference between consecutive terms in AP must be equal.

Alternatively, maybe try to find an arithmetic progression with step r=3.

Check: Starting at 1: 1,4,7,... but 7 isn't in S. Nope.

Starting at 2: 2,5,8,11≡0, not in set.

Wait, maybe starting at a different point?

How about starting at 4:

4,7,… nope.

Alternatively, perhaps r=4?

Start at 1: 1,5,9,13≡2,6,… Hmm, 1,5,9,2,6,10,14≡3. But 3 isn't in S either. So not matching.

Wait, maybe the step is different? r= something else?

Alternatively, perhaps it's a longer step but with wrap-around modulo 11.

But this might be complicated.

Wait, perhaps try another \( a \). Let's take \( a=6 \), which is also a primitive root mod11. Compute its powers:

\(6^0 =1\), \(6^1=6\), \(6^2=36≡3\), \(6^3=18≡7\), \(6^4=42≡9\), \(6^5=54≡10\), \(6^6=60≡5\). So S={1,3,5,6,7,9,10}.

Same as before? No, different set: {1,3,5,6,7,9,10}. Hmm. Let's see if this can form an AP.

Sorted: 1,3,5,6,7,9,10

Differences between terms: 2,2,1,1,2,1.

Not consistent.

Wait, is there any arithmetic progression within these numbers?

If r=2: starting at1:1,3,5,7,9,11≡0. But we don't have 0 or 11, which would be needed for the full AP.

Alternatively, maybe overlapping segments? Not sure.

Wait, perhaps another \( a \). Let me try \( a=7 \) mod11.

Compute powers of 7:

\(7^0 =1\), \(7^1=7\), \(7^2=49≡5\), \(7^3=35≡2\), \(7^4=14≡3\), \(7^5=21≡10\), \(7^6=70≡4\). So S={1,2,3,4,5,7,10}.

Sorted: 1,2,3,4,5,7,10

Differences between consecutive terms:1,1,1,1,2,3.

So nope. Not an AP.

Wait, another primitive root is \( a=8 \).

Compute its powers:

\(8^0 =1\), \(8^1=8\), \(8^2=64≡9\), \(8^3=72≡6\), \(8^4=48≡4\), \(8^5=32≡10\), \(8^6=80≡3\). So S={1,3,4,6,8,9,10}.

Sorted:1,3,4,6,8,9,10. Differences: 2,1,2,2,1,1. Not consistent.

So none of the primitive roots seem to produce a set that can be rearranged into an arithmetic progression with constant difference modulo p=11.

But is this always the case? Maybe for primes where -1 is a quadratic residue or something?

Wait, wait, another idea: Since we have terms \( 1, a, a^2, ..., a^{p-5} \), which are (p-4) terms. So maybe the multiplicative order of a modulo p has to be equal to p-4? But that seems restrictive because for primes p>5, p-4 is not necessarily the order.

Wait, no. The multiplicative group mod p is cyclic with order p-1, so possible orders are divisors of p-1.

But since we have a sequence of length p - 4, perhaps if a has an order such that its powers give us an AP after rearrangement.

Alternatively, maybe when the powers form an additive subgroup? But additive subgroups mod p are cyclic only for prime orders?

Wait, not necessarily. The additive group is always cyclic and isomorphic to Z_p, but multiplicative groups can have different structures.

But I'm getting confused here.

Wait, perhaps another approach: Let me see what condition must hold for such a rearrangement to be possible.

Suppose that the sequence \( b_0, b_1, ..., b_{p-5} \) is an arithmetic progression with difference r modulo p. Then,

\( b_n = b_{n-1} + r \mod p \).

Which means:

\( b_n - r = b_{n-1} \mod p \).

So in terms of the sequence, each term is previous plus r.

Given that, the entire set \( S = \{b_0, b_1, ..., b_{p-5}\} \) must be an arithmetic progression with length \( p -4 \), common difference \( r \).

Therefore, such a set can also be represented as:

\( S = \{c + kr \mod p | k=0, 1, 2, ..., p-5\} \)

for some starting term c.

Thus, the original sequence \(1, a, a^2, ..., a^{p -5}\) must be a permutation of this additive progression set.

So the question is equivalent to: For which primes \( p >5 \), there exists an integer \( a \) such that the multiplicative subgroup generated by \( a \) modulo \( p \) (of size at least \( p -4 \)) contains an arithmetic progression of length \( p-4 \).

Wait, but multiplicative groups are cyclic, so they can't contain long additive progressions unless...

Hmm, I remember from number theory that in a multiplicative group, there aren't generally long additive sequences because addition and multiplication behave differently. Unless the prime is very special.

In particular, for primes where 2 divides p-1, maybe?

Wait, not necessarily.

Alternatively, perhaps only possible when \( p \) is such that \( a \) has a small order so that the powers lie close together in the multiplicative group. But I'm not sure.

Wait, but we're dealing with additive progressions modulo p. So even if the multiplicative subgroup is cyclic, arranging its elements into an arithmetic progression would require some specific conditions on how their logarithms (in the multiplicative sense) are distributed additively.

This seems to relate to something called "additive characters" or maybe exponential sums in number theory.

Alternatively, maybe it's only possible when \( a \) is 1? But if \( a =1 \), all terms would be 1, so can't form an AP with more than one term. So that's not useful.

Wait another thought: If the multiplicative subgroup generated by \( a \) modulo \( p \) has size greater than something, but perhaps here we need it to have size exactly \( p -4 \). Wait, no, since the sequence is up to exponent \( p-5 \), so there are \( p -4 +1 = p -3 \) elements. Hmm.

Wait, hold on: The number of terms is from exponent 0 to exponent \( p -5 \), which gives us \( (p -5) +1 = p -4 \) terms. So we have a set of size \( p -4 \).

But the multiplicative group mod p has order \( p-1 \). So unless \( p -4 \) divides \( p -1 \), but \( p -4 \) divides \( p -1 \) only if 3 divides 1, which is never. Wait, not necessarily; let's compute:

If \( (p -1) / d = k \), where \(d\) is the order of a.

Wait, maybe it's too vague.

Alternatively, perhaps using the concept that in finite fields, additive characters and multiplicative functions don't overlap much except for specific cases. So maybe only certain primes allow this kind of structure.

Let me test another prime: p=13.

So for p=13, we have sequence length p -4 =9 terms. Let's see if there exists an a such that the powers up to a^8 can be arranged into an arithmetic progression with difference r mod13.

First, pick a primitive root modulo13: 2,6,7,11 are primitive roots.

Let me try \( a=2 \). Compute its powers:

2^0 =1

2^1=2

2^2=4

2^3=8

2^4=16≡3

2^5=6

2^6=12

2^7=24≡11

2^8=22≡9

So S={1,2,3,4,6,8,9,11,12}

Now, can we arrange these into an arithmetic progression?

Let me list all elements: 1,2,3,4,6,8,9,11,12.

We need to arrange them in order with a common difference r. Let's see:

Looking at the numbers, perhaps starting from 1 and trying r= something.

Trying r=1:

1,2,3,4,5,... But we don't have 5,7,10. So nope.

r=2:

Starting at 1: 1,3,5,7,9,11,13≡0. Not matching our set.

Wait, our set is {1,2,3,4,6,8,9,11,12}. So let me see if we can arrange them as an AP.

Let's compute all possible differences:

Looking for a starting point and common difference that can cover all elements.

Starting at 1, r=3: 1,4,7,10,... No good.

Starting at 2, r=1: same issue.

Wait, maybe the sequence wraps around? For example, 12,1,2,3,... but then it's not a constant difference.

Alternatively, let me consider all possible pairs and see if there exists an r such that many elements are spaced by r. But with so many terms, it's hard.

But another idea: The number of arithmetic progressions in the multiplicative group would be limited unless some specific conditions hold.

Wait, but maybe for p=13, since 12 is divisible by several numbers, perhaps the set can form an additive progression.

Alternatively, take r=5:

Starting at 1:1,6,11,16≡3,8,13≡0, etc. Not matching.

Wait, our S contains elements {1,2,3,4,6,8,9,11,12}. Let me check if any arithmetic progression with length 9 exists within these numbers mod13.

Is there a starting point and r such that all elements are covered?

Alternatively, perhaps the AP is not necessarily starting at 1 or another term but can start anywhere.

Wait, for example: Let's see if {2,4,6,8,10,12} is an AP with difference 2. But our set includes these even numbers and some odd ones too.

Alternatively, maybe split the progression into two interleaved sequences?

But this might not help.

Wait, actually, given that S contains both small and large residues mod13, it's difficult to see how they can form a linear sequence unless r is 1 or something. But as tried earlier, with r=1, we don't have the necessary elements.

Alternatively, perhaps p=7 was a special case because when p-5=2, which allows for short sequences that are more flexible?

Wait, let's think about how arithmetic and geometric sequences interact here.

In general, an arithmetic progression has terms increasing linearly, while a geometric progression increases multiplicatively. For them to coincide after permutation is very restrictive.

I recall that in finite fields, sometimes multiplicative characters can be converted into additive ones via Gauss sums, but I don't know if that helps directly here.

Wait another angle: The condition given implies that the sequence \( b_n \) must satisfy a linear recurrence relation modulo p. Since each term differs by r from the previous one, it's like a first-order linear recurrence.

But for geometric sequences, they satisfy multiplicative recurrences, which are nonlinear. So perhaps only in specific cases can these two structures overlap?

Alternatively, maybe when \( a \equiv 1 + kr \mod p \) or something similar? But this is vague.

Wait, but considering the problem: The sequence of powers of a must be rearrangeable into an arithmetic progression. That implies that their logarithms (if we consider multiplicative to additive) would form an additive sequence.

But logarithm here would mean discrete log in the multiplicative group. So if I take the discrete logarithm of each term with base a, it should give me an arithmetic progression in exponents. But since exponents are already 0,1,...p-5, maybe this is trivial?

Wait, perhaps not. Wait, the terms are \( a^k \), so their discrete logs to base a would just be k. So rearranging them as exponents doesn't help.

Alternatively, if we have an additive progression in the multiplicative group, it's equivalent to some sort of affine function on the exponents? Hmm, not sure.

Wait, maybe think in terms of generator g of the multiplicative group mod p. Then every element can be written as \( g^m \) for some m. So if our set S is {g^{m_0}, g^{m_1}, ..., g^{m_{p-5}}} }, and we want these to form an additive progression.

But additive progressions involve linear functions in the exponent when expressed multiplicatively, which seems restrictive unless m_i are in arithmetic progression themselves. But since exponents correspond directly, if our set S is {g^k | k = c + dr } for some constants c,d,r, then that might be a possibility. But how likely is this?

In other words, if the exponents form an additive progression, then their images under the generator would form a geometric sequence in multiplicative terms, but since we are dealing with powers of another element \(a\), perhaps only when \(a\) itself is a generator raised to some power that allows such alignment.

This seems too vague. Maybe it's better to look for specific cases where p-4 divides something or the set S can be arranged as AP.

Wait, for p=7, we saw that S={1,3,2} which was an AP.

For p=11, trying multiple primitive roots didn't result in such a set. Similarly for p=13.

Is it possible only when p-4 is small? For example, p=7 gives 3 terms; maybe manageable. Let's check the next prime after 7 where p-5 is reasonable: p=11, which as above doesn't work.

Wait, let's try p=5, but problem specifies p>5.

So perhaps only p=7 works?

But wait, before concluding that, maybe check another small prime. Let me try p=17, just to see the pattern.

For p=17, we have a sequence of length 13 terms (from exponent 0 to 12). So S has 13 elements, which is quite long.

Take \( a =3 \), primitive root mod17.

Compute its powers:

3^0=1

3^1=3

3^2=9

3^3=27≡10

3^4=30≡13

3^5=39≡5

3^6=15

3^7=45≡11

3^8=33≡16

3^9=48≡14

3^{10}=42≡8

3^{11}=24≡7

So S={1,3,5,7,8,9,10,11,13,14,15,16}.

Wait, that's 12 elements. Wait, no: exponent from 0 to 11 is 12 terms? No, wait p-5=12 for p=17, so we have 13 terms (exponents 0-12). So I must compute up to 3^{12}.

Compute 3^12: 3^6=15; 3^7=45≡11; 3^8=33≡16; 3^9=48≡14; 3^{10}=42≡8; 3^{11}=24≡7; 3^{12}=21≡4.

Thus, S={1,3,4,5,7,8,9,10,11,13,14,15,16}.

Now can we arrange these into an arithmetic progression mod17? Hmm, let's see. The elements are:

1,3,4,5,7,8,9,10,11,13,14,15,16.

Looking for a common difference r such that starting from some c, adding r each time modulo17 gives all these numbers in order.

Let me check possible differences:

Difference of 1: Starting at 1, the sequence would be 1,2,... which is missing. But our set has many consecutive numbers except between 1 and3, so maybe not.

Looking for an AP with difference r=2. Let's see:

Starting from1:1,3,5,7,9,11,13,15,17≡0,... Not matching since we don't have 0 or other numbers beyond 16.

But our set is missing 2,6,12. So that doesn't help.

Difference r=3: Starting at1:1,4,7,10,13,16,2,... but again, not matching all elements.

Alternatively, starting from another number with difference 1:

Looking at the set, it contains consecutive numbers except between 1-3 and 11-13. Wait, no, actually, in S we have 1, then skips to3; 4 is present,5 is there,7 is present,8,9,10,11 is missing, wait no:

Wait, the set is {1,3,4,5,7,8,9,10,11,13,14,15,16}.

So 1, then skips to3; but in reality, it's 1,3,4,5,7,8,9,10,11,13,14,15,16. So between 1 and3 is a gap of2.

Then from3 to4:gap1

From4 to5: gap1

From5 to7: gap2

From7 to8: gap1

From8 to9: gap1

From9 to10: gap1

From10 to11: gap1

From11 to13: gap2

From13 to14:gap1

From14 to15:gap1

From15 to16:gap1.

So the gaps alternate between 1 and sometimes 2. It's not consistent, so can't form an AP with fixed r.

Alternatively, perhaps if I arrange the terms in a different order.

Wait, is there a way to interleave them? For example:

Starting at 4, maybe?

But it's getting too vague. Maybe p=17 also doesn't work.

So far, only p=7 works.

Another approach: Maybe considering that for such an AP to exist, the set S must be additively closed or something similar, but in general multiplicative sets aren't additively closed unless specific conditions hold.

I recall from additive combinatorics that sum-product phenomena suggest that sets with both additive and multiplicative structure are rare. So maybe only certain primes allow such a coincidence where the powers of a can form an AP.

Given that p=7 works, but others like 11,13,17 don't seem to work, perhaps p=7 is the only prime satisfying this condition.

Wait, let me check another small prime: p=5. But problem says p>5, so starting from p=7.

Is there a mathematical reason why p=7 works?

For p=7, the multiplicative group has order6, and we're taking 3 terms (p-4=3). So S is {1,a,a^2}, which can be an AP. Indeed, as I saw with a=3, it worked.

But for larger primes, perhaps such a coincidence doesn't happen because you need the multiplicative set to align with additive structure over more elements, which becomes less likely as p increases.

Alternatively, maybe primes where p-1 divides something? Not sure.

Wait, another idea: For an arithmetic progression of length k, we can use the concept from Van der Waerden's theorem or Szemerédi's theorem on arithmetic progressions. But those are more about guaranteeing existence in large enough sets, which is different here.

Alternatively, maybe think about it as a system of equations: The condition \( b_n = b_{n-1} + r \mod p \) for all n implies that:

\( b_1 = b_0 + r \)

\( b_2 = b_0 + 2r \)

...

\( b_{p-5} = b_0 + (p -5)r \).

But since each \( b_i \) is a power of \( a \), say \( a^{k_i} \), then we have:

\( a^{k_n} = a^{k_{n-1}} + r \mod p \)

for all n.

This is similar to the concept of exponential polynomials or linear recurrences with multiplicative terms. It's quite restrictive because each term must satisfy both an additive and multiplicative condition.

Alternatively, in log terms (discrete logarithm), if we let \( a = g^m \) for some generator \( g \) mod p, then the exponents would be multiples of m. But I don't know if that helps here.

Wait, maybe instead consider that since S is an additive progression, then every element except the first can be written as previous plus r. So starting from b0=1 (for simplicity), then:

b1 = 1 + r

b2 = 1 + 2r

...

bk =1 + kr

But all these must be powers of a.

So we have that for each k, \(1 + kr\) is equal to some power of a modulo p. So in other words,

\( a^{m_k} ≡ 1 + kr \mod p \) for each term in the progression.

This resembles solving multiple exponential equations:

\( a^{m_1} ≡ 1 + r \)

\( a^{m_2} ≡ 1 + 2r \)

...

\( a^{m_{p-5}} ≡ 1 + (p -5)r \mod p \).

For this to hold for some integers \( m_k \), it's necessary that each term \(1 + kr\) is in the multiplicative subgroup generated by a modulo p. Since S has size p -4, which must divide the order of the group p-1, but p-4 doesn't necessarily divide p-1 unless specific conditions hold.

Wait, for example, when p=7:

p-1=6 and p-4=3 divides 6, so it's possible. In that case, a can be chosen such that its order is 3, which is exactly what we did with a=3 (since 3^3≡1 mod7).

Similarly, for other primes where p-4 divides p-1, perhaps similar things can happen.

So let me check: For p=7, p-4=3 and p-1=6; 3 divides 6. So that's good.

For p=11: p-4=7, p-1=10. 7 doesn't divide10. Not possible.

p=13: p-4=9, p-1=12; 9 divides12? No, since 12/9=4/3, not integer. So nope.

p=17: p-4=13, p-1=16. Doesn't divide.

Similarly, the next prime where p-4 divides p-1 would be?

Looking for primes where (p -4) divides (p -1). Which implies that (p -4) divides 3, because:

If d divides both p-1 and p-4, then it must divide their difference: (p -1)-(p -4)=3.

Therefore, the possible divisors are 1 and 3. Thus, for (p -4) to divide (p -1), (p -4) must be a divisor of 3.

So:

Case 1: p -4=1 => p=5. But problem specifies p>5.

Case2: p-4=3=> p=7. So the only prime where p -4 divides p -1 is p=7.

Therefore, for all primes p>5 except p=7, (p -4) does not divide (p -1), meaning that such a subgroup of size p -4 doesn't exist in the multiplicative group mod p. Hence, it's impossible to have an arithmetic progression covering exactly these terms because their number doesn't align with any possible subgroup.

Therefore, only p=7 satisfies this condition.

**Final Answer**
The prime is \boxed{7}.
</think>



To solve the problem, we need to find all primes \( p > 5 \) for which there exists an integer \( a \) and an integer \( r \) satisfying \( 1 \leq r \leq p-1 \) such that the sequence \( 1, a, a^2, \ldots, a^{p-5} \) can be rearranged to form a sequence \( b_0, b_1, b_2, \ldots, b_{p-5} \) where \( b_n - b_{n-1} - r \) is divisible by \( p \) for \( 1 \leq n \leq p-5 \).

First, we note that the sequence \( 1, a, a^2, \ldots, a^{p-5} \) must be rearrangeable into an arithmetic progression modulo \( p \). This implies that the set of powers of \( a \) must form an additive sequence with a common difference \( r \).

We checked several primes and found that for \( p = 7 \), we can find such \( a \) and \( r \). Specifically, when \( a = 3 \), the sequence \( 1, 3, 2 \) (which are powers of 3 modulo 7) forms an arithmetic progression with difference \( r = 1 \).

For other primes like 11, 13, and 17, we found that it was not possible to rearrange the powers of any integer \( a \) into such an arithmetic progression. This is because the conditions required for forming such a sequence are very restrictive and only hold when the number of terms (which is \( p - 4 \)) divides the order of the multiplicative group modulo \( p \) (which is \( p - 1 \)).

We concluded that the only prime \( p > 5 \) where \( p - 4 \) divides \( p - 1 \) is \( p = 7 \).

Thus, the final answer is:

\[
\boxed{7}
\]