<Question>
Steve says to Jon, "I am thinking of a polynomial whose roots are all positive integers. The polynomial has the form $P(x) = 2x^3-2ax^2+(a^2-81)x-c$ for some positive integers $a$ and $c$. Can you tell me the values of $a$ and $c$?"

After some calculations, Jon says, "There is more than one such polynomial."

Steve says, "You're right. Here is the value of $a$." He writes down a positive integer and asks, "Can you tell me the value of $c$?"

Jon says, "There are still two possible values of $c$."

Find the sum of the two possible values of $c$.
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem step by step. So, Steve is thinking of a polynomial of the form P(x) = 2x³ - 2a x² + (a² - 81)x - c, where a and c are positive integers, and all the roots are positive integers. Jon initially thinks there's only one such polynomial, but after some calculations, he realizes there are multiple possibilities. Then, after Steve gives the value of a, Jon still can't determine c uniquely, meaning there are two possible values. We need to find the sum of these two c values.

First, let's recall that for a cubic polynomial with roots r, s, t (all positive integers), we can express it using Vieta's formulas. The standard form would be P(x) = k(x - r)(x - s)(x - t), where k is the leading coefficient. In this case, the leading coefficient is 2, so expanding that should match the given polynomial.

<prompt_1.1> Use Vieta's formulas to express the given polynomial in terms of its roots and coefficients. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
Let me write down Vieta's formulas for a cubic polynomial. If the polynomial is 2x³ - 2a x² + (a² - 81)x - c, then when we factor it as 2(x - r)(x - s)(x - t), expanding this should give us the coefficients in terms of the roots.

Expanding 2(x - r)(x - s)(x - t):

First, expand (x - r)(x - s)(x - t):

= x³ - (r + s + t)x² + (rs + rt + st)x - rst

Multiply by 2:

= 2x³ - 2(r + s + t)x² + 2(rs + rt + st)x - 2rst

Now, compare this with the given polynomial:

Given: 2x³ - 2a x² + (a² - 81)x - c

So, equate the coefficients:

1. Coefficient of x³: 2 = 2 (matches)
2. Coefficient of x²: -2(r + s + t) = -2a ⇒ r + s + t = a
3. Coefficient of x: 2(rs + rt + st) = a² - 81
4. Constant term: -2rst = -c ⇒ c = 2rst
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
So, our variables are r, s, t (roots, positive integers), and a, c (positive integers). From Vieta, we have:

1. r + s + t = a
2. rs + rt + st = (a² - 81)/2
3. rst = c/2 ⇒ c = 2rst

So, given that a and c are positive integers, and r, s, t are positive integers, we need to find possible triples (r, s, t) such that the above equations hold.

Jon initially can't determine the polynomial, meaning there are multiple triples (r, s, t) with the same a but different c. Then, after knowing a, Jon still can't determine c, so there are two different c's corresponding to the same a but different triples.

So, the strategy is:

1. Find possible a's such that there are multiple triples (r, s, t) with the same a but different c.
2. For each such a, compute c = 2rst and find the sum of distinct c's.

First, let's work with the equations.

From equations 1 and 2:

rs + rt + st = (a² - 81)/2

But since r + s + t = a, let's denote S = a = r + s + t

Then, rs + rt + st = (S² - 81)/2

But we also know that for three variables:

(r + s + t)² = r² + s² + t² + 2(rs + rt + st)

So, S² = r² + s² + t² + 2(rs + rt + st)

But we can rearrange this to:

r² + s² + t² = S² - 2(rs + rt + st)

Substitute rs + rt + st from above:

= S² - 2*((S² - 81)/2) = S² - (S² - 81) = 81

Therefore, r² + s² + t² = 81

So, now we have:

r + s + t = S = a

rs + rt + st = (S² - 81)/2

r² + s² + t² = 81

So, given that r, s, t are positive integers, their squares sum to 81. Also, their sum is a, which is a positive integer, and their pairwise products sum to (a² - 81)/2.

So, the key equations are:

1. r + s + t = a
2. r² + s² + t² = 81
3. rs + rt + st = (a² - 81)/2

Additionally, since r, s, t are positive integers, each of them is at least 1. So, the smallest possible value for each root is 1.

Given that r² + s² + t² = 81, each of r, s, t must be less than or equal to 9, since 10² = 100 > 81.

So, possible roots are integers from 1 to 9. Let's find all possible triples (r, s, t) with r ≤ s ≤ t (to avoid permutations) such that r² + s² + t² = 81.

<prompt_2.1> Find all possible triples (r, s, t) with r ≤ s ≤ t such that r² + s² + t² = 81. </prompt_2.1>
</Plan_2>

<Execution_2>
<execution_2.1>
Let me list them.

First, start with the largest possible t. Since t is the largest, let's start with t = 9.

Case 1: t = 9

Then, r² + s² = 81 - 81 = 0. But r and s are positive integers, so this is impossible. So, t cannot be 9.

Case 2: t = 8

Then, r² + s² = 81 - 64 = 17

Find r, s ≤ 8, r ≤ s, such that r² + s² = 17.

Possible pairs:

1² + 4² = 1 + 16 = 17

So, (1, 4, 8)

Check if there are others:

2² + 3² = 4 + 9 = 13 ≠17

So, only (1,4,8)

Case 3: t = 7

Then, r² + s² = 81 - 49 = 32

Find r, s ≤7, r ≤ s, such that r² + s² =32

Possible pairs:

4² + 4² = 16 +16=32

So, (4,4,7)

Also, check others:

5² + something? 5²=25, 32-25=7, which is not a square.

3²=9, 32-9=23, not a square.

2²=4, 32-4=28, not a square.

1²=1, 32-1=31, not a square.

So only (4,4,7)

Case 4: t=6

Then, r² + s²=81 -36=45

Find r, s ≤6, r ≤s, such that r² +s²=45

Possible pairs:

3² + 6²=9 +36=45

So, (3,6,6)

Also, 5² + something? 25 +20=45, 20 not square.

4²=16, 45-16=29, not square.

2²=4, 45-4=41, nope.

1²=1, 45-1=44, nope.

So, only (3,6,6)

Case5: t=5

Then, r² +s²=81 -25=56

Find r, s ≤5, r ≤s, such that r² +s²=56

Possible pairs:

5² + something? 25 +31=56, nope.

4²=16, 56-16=40, not square.

3²=9, 56-9=47, nope.

2²=4, 56-4=52, nope.

1²=1, 56-1=55, nope.

So, no solutions here.

Case6: t=4

Then, r² +s²=81 -16=65

But since t=4 and r ≤s ≤t=4, so r and s can be at most 4.

Possible pairs:

4² + something? 16 +49=65, but 49 is 7² which is more than 4.

3²=9, 65-9=56, which is 7.48, not integer.

Similarly, 2²=4, 65-4=61, nope.

1²=1, 65-1=64=8², but 8>4. So, no solutions.

Case7: t=3

Then, r² +s²=81 -9=72

But r and s ≤3. 3² +3²=18, which is way less than 72. So, impossible.

Similarly, t=2 or t=1 would result in even larger required sums for r and s, which is impossible. So, the only possible triples are:

(1,4,8), (4,4,7), (3,6,6)

But wait, we need to check all permutations? Wait, no, since we considered r ≤s ≤t, so these are the only three distinct unordered triples.
</execution_2.1>
</Execution_2>

<Plan_3>
Based on execution_2.1:
Let's verify each:

For (1,4,8):

Sum: 1 + 4 + 8 = 13

Sum of squares: 1 + 16 + 64 = 81

Sum of products: (1*4 + 1*8 + 4*8) = 4 + 8 + 32 = 44

Then, a = 13, and c = 2*1*4*8 = 64

For (4,4,7):

Sum: 4 + 4 + 7 = 15

Sum of squares: 16 + 16 + 49 = 81

Sum of products: (4*4 + 4*7 + 4*7) = 16 + 28 + 28 = 72

Then, a =15, c=2*4*4*7=112

For (3,6,6):

Sum: 3 +6 +6=15

Sum of squares:9 +36 +36=81

Sum of products: (3*6 +3*6 +6*6)=18 +18 +36=72

So, same as (4,4,7), a=15, c=112

Wait, so two different triples with same a=15 and same c=112.

So, Jon first thought there was only one polynomial, but actually for a=15, there are two triples (4,4,7) and (3,6,6), both giving c=112. Wait, but hold on, (3,6,6) gives the same c as (4,4,7). So, maybe Jon thought there was only one polynomial, but actually, for a=15, there are two polynomials with same a but same c? But the problem says Jon says there's more than one polynomial, so maybe a is not unique.

Wait, but let's check.

Wait, Jon initially says, "more than one such polynomial," so there must be multiple a's? Wait, no.

Wait, the polynomial is determined by a and c. So, if two different triples (r,s,t) give the same a but different c, then they would be different polynomials. But in our case, for a=15, both triples give the same c=112, so the polynomial would be the same. So, maybe Jon was referring to multiple a's?

Wait, maybe Jon was considering different a's. Wait, let's check.

Wait, let's see all possible triples:

We found three triples:

1. (1,4,8): a=13, c=64

2. (4,4,7): a=15, c=112

3. (3,6,6): a=15, c=112

So, a=13 and a=15.

So, if a=13, only one c=64.

If a=15, two c's: 112 and maybe another one? Wait, but in our list, only two triples for a=15.

Wait, but wait, are there more triples?

Wait, maybe we missed some. Let's check again.

<prompt_3.1> Verify if there are any other possible triples (r, s, t) that satisfy the conditions. </prompt_3.1>
-----

So, the answer is 112 + 128 = 240.

But let me check once more if there are other possible triples.

<prompt_3.2> Verify if there are any other possible triples (r, s, t) and conclude the solution. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, when t=8, we had (1,4,8). Are there other combinations for t=8?

r² + s²=17. We found (1,4). Are there others?

s=4, r=1. s= something else?

For example, r=2, s²=17 -4=13, not square. r=3, s²=17 -9=8, not square. So, only (1,4,8).

For t=7, we had (4,4,7). Any others?

r² + s²=32. We found (4,4). If r=2, s²=32 -4=28, not square. r=1, s²=31, nope. So, only (4,4,7).

For t=6, (3,6,6). Any others?

r² + s²=45. We found (3,6). If r=5, s²=45 -25=20, not square. r=4, s²=45 -16=29, nope. So, only (3,6,6).

t=5: no solutions.

t=4: no.

t=3: no.

t=2: no.

t=1: no.

So, only three triples. Therefore, only a=13 and a=15. For a=13, c=64; for a=15, c=112 (from two different triples). So, when Jon was told a, if a=15, he would still have two possible c's. If a=13, only one c. But the problem states that after Steve gives a, Jon still can't determine c, meaning there are two possible c's. Therefore, a must be 15, leading to c=112 and 128? Wait, wait, hold on. Wait, in our calculation, both triples for a=15 give c=112. Wait, that can't be. Wait, (4,4,7): rst=4*4*7=112, so c=2*112=224? Wait, wait, wait, hold on. Wait, in the problem statement, c=2rst. Wait, Vieta's formula says the constant term is -rst, so in the polynomial, it's -c, so -rst = -c ⇒ c = rst. Wait, but in the problem statement, the polynomial is given as 2x³ -2a x² + (a² -81)x -c. So, the constant term is -c, which equals -rst (from expansion). Therefore, -c = -rst ⇒ c = rst. But in our earlier calculation, we had c=2rst. Wait, that's a mistake.

Wait, let's go back.

Original polynomial: 2x³ -2a x² + (a² -81)x -c

Compare to 2(x - r)(x - s)(x - t) = 2x³ -2(r + s + t)x² + 2(rs + rt + st)x -2rst

Therefore, equate coefficients:

- Coefficient of x²: -2a = -2(r + s + t) ⇒ a = r + s + t

- Coefficient of x: 2(rs + rt + st) = a² -81 ⇒ rs + rt + st = (a² -81)/2

- Constant term: -2rst = -c ⇒ c = 2rst

Therefore, c = 2rst. So, in the previous calculation, c is twice the product of the roots. Therefore, for the triple (4,4,7):

rst = 4*4*7 = 112 ⇒ c = 2*112 = 224

But earlier, I thought c=112, but that's incorrect. Similarly, for (3,6,6): rst=3*6*6=108 ⇒ c=2*108=216

Wait, so that's a mistake. Let me correct that.

So, for each triple:

1. (1,4,8):

rst =1*4*8=32 ⇒ c=64

2. (4,4,7):

rst=4*4*7=112 ⇒ c=224

3. (3,6,6):

rst=3*6*6=108 ⇒ c=216

Therefore, the possible c's are 64, 224, 216.

But Jon says there are more than one polynomial, meaning multiple possible triples. Then, when Steve gives a, if a=13, then only one c=64. If a=15, two c's: 224 and 216. Therefore, Jon, after being told a=15, would see two possible c's, hence says there are two possible c's. Therefore, the answer is 224 + 216 = 440.

Wait, but let me check again.

Wait, the problem says: "Jon says, 'There is more than one such polynomial.' Steve says, 'You're right. Here is the value of a.' Jon says, 'There are still two possible values of c'."

So, initially, Jon couldn't determine c, so there must be multiple possible c's for the same a. Then, after knowing a, Jon still can't determine c, meaning that for that a, there are multiple c's. So, the possible a's are those that have multiple c's. So, in our case, a=15 has two c's (224 and 216). Therefore, Jon, upon being told a=15, would see two c's. Therefore, the sum is 224 + 216 = 440.

But let's confirm if there are any other a's.
</execution_3.1>

<execution_3.2>
Wait, in the case of t=6, we had (3,6,6). Are there other permutations? For example, (6,6,3) but since we considered r ≤s ≤t, it's the same as (3,6,6). Similarly for others.

Wait, but let's check if there are other triples with different numbers but same sum of squares. For example, maybe (2,5,8). Let's check sum of squares: 4 +25 +64=93≠81. So no.

Or (2,6,7): 4 +36 +49=89≠81. No.

Wait, maybe (5,5,5): 25*3=75≠81. No.

Or (2,2,9): sum of squares 4 +4 +81=89≠81. No.

Wait, (1,1, sqrt(79)): but sqrt(79) is not integer. So, no.

So, seems like only three triples. So, a=13 and a=15. For a=13, c=64. For a=15, c=112. So, Jon first says multiple polynomials, so a must be 15 (since a=13 only gives one c). Then, Steve tells a=15, and Jon still can't determine c, meaning there are two possible c's. Wait, but in our case, both triples for a=15 give c=112. So, that would mean only one c. But the problem says Jon says, "There are still two possible values of c". So, that suggests that after knowing a=15, there are two c's. Therefore, maybe there is another triple for a=15 with a different c.

Wait, perhaps we missed a triple. Let me check again.

Wait, maybe there is another triple with sum of squares 81.

Wait, let's check all possible combinations where r ≤s ≤t and r² +s² +t²=81.

We have:

1. 1,4,8: 1+16+64=81

2. 4,4,7: 16+16+49=81

3. 3,6,6:9+36+36=81

Is there another one?

Let's check t=5:

r² +s²=56. Let's see if there are r ≤s ≤5 such that r² +s²=56.

s can be at most 5.

Possible s=5: r²=56 -25=31, not square.

s=4: r²=56 -16=40, not square.

s=3: r²=56 -9=47, no.

s=2: r²=56 -4=52, no.

s=1: r²=56 -1=55, no.

t=7:

Wait, we already did t=7.

Wait, maybe t=5, but r and s can be higher? Wait, no, since t is the largest.

Alternatively, maybe other combinations. Let's try t=5, but with different r and s.

Wait, we already checked. No.

How about t=2:

r² +s²=81 -4=77. But r and s ≤2, max sum of squares 4 +4=8. Not possible.

t=1: same issue.

How about t=8:

Wait, we considered t=8, which gives r=1, s=4, t=8. Any others?

Wait, t=8, r² +s²=17. We found (1,4). Are there others?

s=4: 16 +1=17. s=3: 9 +8=17, but 8 isn't square. s=2:4 +13=17, nope. s=1:1 +16=17. So, only (1,4,8).

t=7: (4,4,7). Any others?

r² +s²=32. We found (4,4). Are there others?

s=5: 25 +7=32, 7 not square.

s=6: 36 + (-4)=32, nope.

s=4: 16 +16=32. So only (4,4,7).

t=6: (3,6,6). Any others?

r² +s²=45. We found (3,6). Any others?

s=6: 36 +9=45. s=5:25 +20=45, nope. s=4:16 +29=45, nope. So only (3,6,6).

So, only three triples.

Therefore, for a=15, both triples give c=112. Therefore, there is only one c. But the problem states that after Steve gives a=15, Jon still can't determine c, implying two possible c's. Therefore, there must be another triple with a=15 and different c. Therefore, perhaps we missed a triple.

Alternatively, maybe there's a case where even though the roots are different, c is the same. Let me check.

Wait, for (4,4,7): rst=4*4*7=112, so c=224? Wait, wait, hold on. Wait, earlier I thought c=2rst. Let me check again.
</execution_3.2>
</Execution_3>

<Plan_4>
Original polynomial: 2x³ -2a x² + (a² -81)x -c

Expanded form: 2(x - r)(x - s)(x - t) = 2x³ -2(r + s + t)x² + 2(rs + rt + st)x -2rst

Therefore, the constant term is -2rst, which equals -c. Therefore, c = 2rst.

Yes, so c is twice the product of the roots. So, for (4,4,7), c=2*4*4*7=224. For (3,6,6), c=2*3*6*6=216. So, two different c's.

Therefore, when a=15, there are two possible c's: 216 and 224. Therefore, Jon, after being told a=15, would see c=216 and c=224. Therefore, the sum is 216 + 224 = 440.

But let me check if there are other possible triples. Wait, maybe there is a case where two different triples give the same a and same c. For example, if two different triples (r,s,t) and (p,q,w) give same a and same c. But in our case, a=15, c=216 and c=224. So, different c's. So, Jon, upon knowing a=15, can't determine c because c can be 216 or 224. Therefore, the answer is 216 + 224 = 440.

But let me confirm once more if there are other possible triples. For example, let's check if there is another triple with sum of squares 81.

Wait, maybe (2, 5, 8):

2² +5² +8²=4 +25 +64=93≠81.

Or (2,7,6): same as (2,6,7), sum of squares 4 +36 +49=89≠81.

(3,4,8):9 +16 +64=89≠81.

(5,5,5):75≠81.

(1,5,9):1 +25 +81=107≠81.

(1,6,8):1 +36 +64=101≠81.

(2,4,9):4 +16 +81=101≠81.

(2,3,9):4 +9 +81=94≠81.

(3,3,9):9 +9 +81=99≠81.

(4,5,8):16 +25 +64=105≠81.

(5,6,7):25 +36 +49=110≠81.

So, no other triples. Therefore, only three triples, leading to a=13,15,15 with c=64,224,216. Therefore, Jon, after knowing a=15, can't determine c because there are two possibilities:216 and 224. Therefore, the sum is 216 +224=440.
</Plan_4>

<Execution_4>
No prompt to execute.
</Execution_4>

<Final_answer>
Therefore, the sum of the two possible values of $c$ is $\boxed{440}$.
</Final_answer>
</think>