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Topic: A non-zero exponent puzzle (Read 682 times) |
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K Sengupta
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A non-zero exponent puzzle
« on: Jan 21st, 2007, 8:20am » |
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Determine analytically all possible non-zero integer solutions to this equation:
p^3 = q(q^8 - 9*q^5+ 81q- 81)
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Eigenray
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Re: A non-zero exponent puzzle
« Reply #1 on: Jan 21st, 2007, 9:44pm » |
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I know there are only finitely many solutions. There are four cases to consider, and the four relevent equations are x3-y3=1,3, whose solutions are evident; and x3-2y3=1,3, whose only solutions seem to be x=1,-1,-5, but in any case are finite in number by Thue's theorem.
So presumably (p,q)=(-2,1) is the only non-zero solution.
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Eigenray
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Re: A non-zero exponent puzzle
« Reply #2 on: Jan 22nd, 2007, 9:23am » |
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Okay, I have an elementary proof:
| hidden: | p3 = q(q2-3)(q2+3)(q4-9q+9) =: q*A*B*C.
1) First suppose q is not divisible by 3. Then none of A,B,C are divisible by 3. Moreover, q is relatively prime to A,B,C, and so q=r3 and ABC must both be cubes.
Note C = AB - 9(q-2). Since A = (q-2)(q+2)+1, it follows gcd(A,C)=1. Also, B = (q-2)(q+2)+7, so (B,C) is either 1 or 7. But q=r3 is either 0,1, or 6 mod 7, so B is not divisible by 7. Hence (B,C)=1.
Now, AB is relatively prime to C, so AB=s3 is a cube, and 9 = q4-AB = (r4)3-s3, and the only way to write 9 as a difference of cubes is 8 - (-1) or 1 - (-8). So r4=2 (impossible), or r4=1, which gives r=+/-1 and q=+/-1. q=1 gives p3=-8, okay, but q=-1 gives p3=152, no good. So we get one solution: q=1 (and p=-2).
2) Now suppose q=3t. We have
p3 = 35t(3t2-1)(3t2+1)(9t4-3t+1) =: 35t*X*Y*Z.
Now X,Y,Z are relatively prime to 3 and t, so 35t and XYZ are both cubes. Write t=3r3.
Since Y-X=2, we have (X,Y)=1 (if t is even) or 2 (if t is odd), and as before we find (X,Z)=1.
Case I: (X,Y)=1. Then X is relatively prime to YZ, so X=s3 is a cube, and 1 = 3t2-X = (3r2)3-s3, and the only cubes differing by 1 are 0 and +/-1. So 3r2 = 1 (impossible) or 0 (giving q=0).
Case II: (X,Y)=2. Then t is odd, so X=2 mod 4, so X/2 is odd, hence relatively prime to Y. Then X/2 is relatively prime to 2YZ, so X/2 is a cube, hence 0,1, or 8 mod 9. But 3|t, so X=8 mod 9, a contradiction. |
It works out pretty nicely after all (I originally had 8 cases to consider).
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| « Last Edit: Jan 22nd, 2007, 9:23am by Eigenray » |
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K Sengupta
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Re: A non-zero exponent puzzle
« Reply #3 on: Jan 23rd, 2007, 6:59am » |
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On Jan 22nd, 2007, 9:23am; Eigenray wrote:
Quote:| (p,q)=(-2,1) is the only non-zero solution |
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True.
However, the problem under reference can also be solved in the following manner.
p^3 = q^9 - 9*q^6 + 81*q^2 -81q
Or, p^3 = (q^3 -3)^3 - (3q-3)^3
By Fermat's Last Theorem, for non-zero integers a, b and c; a^3 cannot equal b^3+c^3. So, relaxing the non-zero restriction
on a,b and c, we observe that:
(a,b,c) = (0, k,-k); (m,0,m); (n,n,0) gives all possible integer solutions for fixed k,m,n....(B)
Now, from (B) we observe that:
(a,b,c) =(0,k,-k) would imply that p=0. This is a contradiction.
(a,b,c) =(m,0,m) would imply that q^3 = 3, so that
q = cuberoot(3), which is not an integer, and thus constitutes a violation of the tenets comprising the problem under reference.
(a,b,c) = (n,n,0) gives 3x=3, so that q=1, giving :
p= 1-3 = -2.
Thus, (q,p) = (1,-2) is the only possible solution to the given problem.
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