Finding Integer Solutions to Polynomial Equations: Can it be Done Easily?

[MiddleEast]

Homework Statement

Is there a number that is one more than its cube?

Relevant Equations

IVT, Graph & proof.

Hello,
Am re-studying math & calculus aiming to start pure math studying later.
However, I got this problem in Stewart calculus.

Typically, this is a straightforward IVT application.
x = x^3 + 1, call f(x)= x^3 - x + 1 & apply IVT.
However I have two things to discuss. First thing is simple proof method & whether it is accepted or not ?
Second thing how to know if there is an integer solution of P(x) = 0 where P is polynomial

Frist : Basically solution of f(x) = g(x) is the x-coordinates of intersection points. I will just draw both curves and point out there is an intersection hence yes there is a number that is 1 more its cube. [Can you accept this a solution if you are a professor?]. Note that question did not ask to use IVT & no exact value is needed.

Second : Another simple way is to find that number! So basically solving x^3 - x + 1 = 0. Here is my question, I solved it online & real solution is ugly. But I remember from high school there is some theorem saying solution is factors of constant term OR coefficient of higher power x. Both of constant term & coefficient is 1 so factors are 1 & -1 but none of them satisfy the equation! I think am missing something or there is some condition in order for this to be true.
TL;DR : Is there any way to know that a polynomial equation has integer solution without solving it?

 

[Orodruin]

Homework Statement:: Is there a number that is one more its cube?
Relevant Equations:: IVT, Graph & proof.

Frist : Basically solution of f(x) = g(x) is the x-coordinates of intersection points. I will just draw both curves and point out there is an intersection hence yes there is a number that is 1 more its cube. [Can you accept this a solution if you are a professor?]. Note that question did not ask to use IVT & no exact value is needed.

No, as a physics professor I might accept a graphical solution but a maths professor would likely want you to formalize that.

 

[topsquark]

TL;DR : Is there any way to know that a polynomial equation has integer solution without solving it?

See here for the Rational Root Theorem.

If you have to solve this one by hand you are going to have to use something like Cardano's method.

-Dan

 

[FactChecker]

Frist : Basically solution of f(x) = g(x) is the x-coordinates of intersection points. I will just draw both curves and point out there is an intersection hence yes there is a number that is 1 more its cube. [Can you accept this a solution if you are a professor?]. Note that question did not ask to use IVT & no exact value is needed.

Are you assuming that there is a clear intersection where f(x) goes from one side of g(x) to the other? If not, they might be tangent and touch, or they might just come so close that you can not see the separation on your graph. If f(x) does go from one side to the other, then apply the IVT to f(x)-g(x). The IVT is a good thing, it is not something to be avoided. Use the tools available.

Would you say that these curves touch at x=0? They don't. The blue curve remains at least 0.00000001 below the green curve.

 

[MiddleEast]

Are you assuming that there is a clear intersection where f(x) goes from one side of g(x) to the other? If not, they might be tangent and touch, or they might just come so close that you can not see the separation on your graph. If f(x) does go from one side to the other, then apply the IVT to f(x)-g(x). The IVT is a good thing, it is not something to be avoided. Use the tools available.

Would you say that these curves touch at x=0? They don't. The blue curve remains at least 0.00000001 below the green curve.
View attachment 314143

Thanks.
Am not avoiding, am thinking in other ways. For this particular problem, surely x intersect with x^3 + 1 if draw both graphs. Clear intersection is there & I understood your point, but for this problem there is clear intersection.

 

[FactChecker]

Thanks.
Am not avoiding, am thinking in other ways. For this particular problem, surely x intersect with x^3 + 1 if draw both graphs. Clear intersection is there & I understood your point, but for this problem there is clear intersection.

Are you "thinking in other ways" or just avoiding saying it by being casual? IMO, you are being casual and just trying not to specifically reference the IVT. But the IVT is really what is making it work.

 

[WWGD]

You can do some basic analysis :
1)If the expression/polynomial is of odd degree, it will have at least one Real zero

2) its not going to be right of $x=1$. For$$x=1$$ , we get 1 on the left,$1^3+1=2$, and g$x^3$ will grow much faster than $x$. Similar for $x=-1$. But you can do the standard analysis using the derivative.

 

[PeroK]

Thanks.
Am not avoiding, am thinking in other ways. For this particular problem, surely x intersect with x^3 + 1 if draw both graphs. Clear intersection is there & I understood your point, but for this problem there is clear intersection.

Yes, but formally you need the intermediate value theorem for continuous functions.

 

[pbuk]

Clear intersection is there & I understood your point, but for this problem there is clear intersection.

Can you accept this a solution if you are a professor?

The answer to this question is no: looking at a picture is never a proof in mathematics. Correct use of the IVT would be an acceptable solution. Nothing more to say*, accept this point and move on.

* Edit: except of course that there is no integer solution. Proving this takes a little more effort and drawing a graph may help you work out how to do it, but the graph itself is still not any part of the proof.

 

[mathwonk]

in a sense, using the IVT rigorously is even easier than graphing. I.e. for graphing you must plot some points, i.e. evaluate the polynomial at some inputs. For IVT all you have to do is find two inputs that give outputs of opposite sign and you are done by the IVT. Since P(0) = 1 and P(-2) = -5, The IVT says that some x with -2 < x < 0, gives P(-2) = 0 . Where of course P(x) = x^3 - x +1.

I.e. for the IVT you only need to plug in two appropriate inputs and evaluate P at them. You don't need also to plot them! So in a sense, (making and) looking at a graph does give you enough to apply the IVT, if the appropriate points (one above and one below the x axis) are correctly marked on the graph. But you do have to put the essential facts into words.