How would Fermat solve this?
2 x 3 + 3 x 2 + 3 x + 1 = x 3 + 6 x 2 + 1 2 x + 8
Find the number of positive integer solutions that satisfy the equation above.
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5 solutions
This is a marvellous solution to the problem and luckily it can be contained in the margin this time.
Correeeeeeeeect!
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Yaaaaay! I solved it :P
@Pi Han Goh I recommend to change the title as it gives away the solution.
Nice problem boi!!
Done it the same way.
What is Fermat's Last theorem do you mean ?
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I didn't recognize the sum of cubes, but by combining terms and factoring I had x(x^2-3x-9)=7. Since 7 is prime and the question asks for a positive integer root the only possible roots are 7 and 1 since the problem is now in the form a*b=p. Neither 7 nor 1 satisfy the equation so there are no positive integer solutions. You can go further and test -1 and -7 and see that there are no integer solutions.
Moderator note:
We could directly apply the integer root theorem, since we have the first and last coefficients.
Right! You applied rational root theorem .
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I solved it by simplifying the expression as ( x + 1 ) 3 = 6 x 2 + 1 2 x + 8 Since R.H.S. is unfactorable in real roots so this equation has no real roots.Is this method correct please correct me if I am wrong.
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The equation you gave is not true. So your reasoning is incorrect. Try simplifying the entire equation first.
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@Pi Han Goh – Okay I understood but if in some other equation this equlity holds is this reasoning correct.I was lucky in this question though.
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@Chaitnya Shrivastava – That depends on what equation you're referring to. There are infinitely many equations out there!
@Pi Han Goh – Substracting x^3 from both sides gives the equation he wrote.
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@Gal Cohen – Yes that's what I was asking to him
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@Chaitnya Shrivastava – Alright. I'll concede to that. But just because RHS has no real roots, that doesn't make the entire equation to have no real roots.
For example, the RHS of x 3 + 3 x − 1 = − 3 x 2 − 2 has no real roots. But the entire equation has at least one real root.
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@Pi Han Goh – Yes that's true but in my case LHS is a perfect cube so doesn't that make a difference.
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@Chaitnya Shrivastava – Unfortunately, you're still mistaken. A third degree polynomial with real coefficients always has at least 1 real root.
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@Pi Han Goh – Oh! Thank you(just a doubt is there any way to find roots of cubic equation other than rational roots theorem)
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@Chaitnya Shrivastava – There's other ways like numerical approximation. The simplest one to learn is the bisection method .
And another method, which is ridiculously painful is Cardano's method .
I liked the method you did it Owen. It is better than adding another theorem to memory, which I had never heard of earlier.....
The equation can be written in the standard form f ( x ) = x 3 − 3 x 2 − 9 x − 7 = 0 . . By the Rational Root Theorem, any rational root of f ( x ) is among +/-1 and +/-7. But since f(1) = -18, f(-1) = -2, f(7) = 126, and f(-7) = -434, f ( x ) has no rational root, or integral root. The answer is 0.
Correct. In fact, you can just show that f ( 1 ) , f ( 7 ) = 0 because we're only checking positive integer solutions.
Taking it mod x , we see 1 ≡ 8 ( m o d x ) . For positive integers, this can only be true for x = 1 , 7 , but neither of those work.
Transforming it into a cubic equation and upon solving for its roots, no integral value applies for both sides of the equation.
That's right too. But it requires more work. +1
Re-write the equation as x 3 + ( x + 1 ) 3 = ( x + 2 ) 3
Now, by Fermat's Last theorem the given equation has no solution in positive integers .
Thus no value of x satisfies the given equation.