(Original post by RDKGames)
I'm not good at proofs of this sort, especially with rationaility involved, but I had a good midnight go at it and will look into the second part more tomorrow, will update here whenever I figure it outSpoiler:Show
Assume is an integer solution to the above cubic.
Therefore
In the event is even; then and are always even. However, is even, and 291 is odd, therefore is always odd. therefore we reach a contradiction.
In the event is odd; then and are always odd. However, is odd, and 291 is odd, therefore is always even. therefore we reach a contradiction yet again.
Thus we prove there are no integer solutions by means of contradiction. QED.
...and I'm dry on this one for nowIf is rational then it can be expressed in the form .(Original post by RDKGames)
I'm not good at proofs of this sort, especially with rationaility involved, but I had a good midnight go at it and will look into the second part more tomorrow, will update here whenever I figure it outSpoiler:Show
Assume is an integer solution to the above cubic.
Therefore
In the event is even; then and are always even. However, is even, and 291 is odd, therefore is always odd. therefore we reach a contradiction.
In the event is odd; then and are always odd. However, is odd, and 291 is odd, therefore is always even. therefore we reach a contradiction yet again.
Thus we prove there are no integer solutions by means of contradiction. QED.
...and I'm dry on this one for now
Then and so clearly .
Considering cases and changing the parity of a and b leads to LHS being odd but RHS is even so it's clearly not true and thus there is a contradiction. The only case that appears to work is where a and b are both even, but we assumed that a and b were co prime, so it leads to a contradiction again. Therefore no rational solutions.
Year 13 Maths Help Thread

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 28072016 01:32

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 28072016 01:47
(Original post by B_9710)
If is rational then it can be expressed in the form .
Then and so clearly .
Considering cases and changing the parity of a and b leads to LHS being odd but RHS is even so it's clearly not true and thus there is a contradiction. The only case that appears to work is where a and b are both even, but we assumed that a and b were co prime, so it leads to a contradiction again. Therefore no rational solutions. 
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 28072016 10:25
(Original post by Ano123)
[Note. No credit is given for using the rational roots theorem.] 
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 28072016 11:10
(Original post by Zacken)
Why not? 
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 28072016 12:06

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 28072016 12:42
But if you want to find say then it can be done. 
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 28072016 15:50
If , then .
, then .
I have no idea is there is a more 'proper' answer to this but this is what I came up with and verified it a few times.
Of course from what I got you could easily have and you would get 4 answers for the 4 different cases that arise according to the values relating to m and n. 
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 30072016 01:02
Another question here, a supposed Diophantine equation, the question is
Prove that the equation has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.
ALSO
Prove that the equation has no non trivial rational solutions.Last edited by Ano123; 30072016 at 01:08. 
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 30072016 01:07
(Original post by Ano123)
Another question here, a supposed Diophantine equation, the question is
Prove that the equation has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions. 
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 30072016 02:05
(Original post by Ano123)
Another question here, a supposed Diophantine equation, the question is
Prove that the equation has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.
ALSO
Prove that the equation has no non trivial rational solutions.
Posted from TSR Mobile 
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 30072016 02:13
(Original post by Ano123)
Another question here, a supposed Diophantine equation, the question is
Prove that the equation has no integer solutions (besides the trivial solution).
Deduce that the equation also has no rational solutions.
Translated (all credits go to the individuals who wrote the document http://lespascals.org/docs/TrainingMaths.pdf ).
The problem is on the first page.
Spoiler, full solution:Spoiler:ShowThe infinite descent is the principle by which there are no strictly decreasing series of positive integers.
We use this principle to prove that there are no solutions to certain problems involving integers: if based on a solution, we are able to fabricate another one stricly smaller but still in integers, then we can repeat this indefinitely, so the initial problem has no solution.
Exercice: Solve in integers the equation:
Solution:
Suppose that is a solution to the equation, , and being integers.
Then:
So is even.
We pose , an integer such that
Then:
So:
Which can be rewritten as:
So is even. We pose , an integer such that
Then:
So:
Which can be rewritten as:
So is even. We pose , an integer such that
Then:
Which can be rewritten as:
We have deduced from the solution a new solution , stricly smaller than the first.
Thus, we have constructed an infinite serie of solutions entirely decreasing, which is contradictory.
The equation has no solution.
Last edited by MartyO; 31072016 at 22:53. 
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 30072016 11:10

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 30072016 11:33
(Original post by RDKGames)
That's what I thought originally but I think FLT's easy way would only apply if the coefficients were all 1.
Posted from TSR Mobile 
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 30072016 17:30
Self teaching myself FP3 so I'll post anything I don't understand here. Firstly, show that but I keep getting 1/6 rather than +1/6 for the third term. Where am I going wrong here?

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 30072016 17:47
(Original post by RDKGames)
Self teaching myself FP3 so I'll post anything I don't understand here. Firstly, show that but I keep getting 1/6 rather than +1/6 for the third term. Where am I going wrong here?
The expansion of is valid for , but perhaps.Last edited by NotNotBatman; 30072016 at 18:25. 
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 30072016 19:56
(Original post by NotNotBatman)
You need to square the maclaurins expansion of
The expansion of is valid for , but perhaps.

Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x^{3}ln(x) goes to 0 as well as x^{3} so it only makes sense?
Last edited by RDKGames; 30072016 at 19:58. 
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 30072016 20:01
(Original post by RDKGames)
Thanks, I got it in the end.

Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x^{3}ln(x) goes to 0 as well as x^{3} so it only makes sense?
So its essentially 1/0 which is infinity.Last edited by Math12345; 30072016 at 20:02. 
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 30072016 20:06
(Original post by Math12345)
x^k ln x tends to 0 when x tends to 0+.
So its essentially 1/0 which is infinity. 
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 30072016 20:10
(Original post by RDKGames)
Thanks, I got it in the end.

Q3: Surely that should say that it tends to positive infinity because the denominator tends to 0 as x goes to 0? x^{3}ln(x) goes to 0 as well as x^{3} so it only makes sense?

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 30072016 21:55
(Original post by RDKGames)
...Last edited by Zacken; 30072016 at 23:00.
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