# Rational roots of 4th degree polynomial with odd coefficents

1. Aug 24, 2016

### Hiero

1. The problem statement, all variables and given/known data
A polynomial, P(x), is fourth degree and has all odd-integer coefficients. What is the maximum possible number of rational solutions to P(x)=0?

2. Relevant equations
P(x) = k(x-r1)(x-r2)(x-r3)(x-r4)
P(x) = 0 when x = {r1, r2, r3, r4}

3. The attempt at a solution
I expanded the "relevant equation." This maybe isn't a useful approach, but it's the only thing I could think of to try to get any information. This is what I got,

P(x) = kx4 - k(r1+r2+r3+r4)x3 + k(r1r2+r1r3+r1r4+r2r3+r2r4+r3r4)x2 - k(r1r2(r3+r4)+r3r4(r1+r2))x + kr1r2r3r4

So then we must have,

k = an odd integer
kr1r2r3r4 = an odd integer
k(r1r2(r3+r4)+r3r4(r1+r2)) = an odd integer
k(r1r2+r1r3+r1r4+r2r3+r2r4+r3r4) = an odd integer

And the goal is to find out the maximum number of {r1, r2, r3, r4} which can be rational.

I really have no idea how to figure this problem. Any help is appreciated.

2. Aug 24, 2016

Maybe I'm wrong, but I think the answer is 4. Try picking 4 rational roots at random and see if you don't generate a polynomial with odd coeficients. One suggestion, for the 4 rational roots, choose fractions with odd integers. You can't get more than 4, but my guess is that you can get 4. editing... I think the problem is harder than it looks. A little work expanding the factors with rational roots shows it doesn't appear you can get all odd coefficients. It leaves me at the drawing board=I don't have an answer.

Last edited: Aug 24, 2016
3. Aug 24, 2016

### Hiero

In case it helps anyone to solve the problem, the answer is actually zero.

I'm very curious about the solution so please explain or hint at it if you work it out.

4. Aug 24, 2016

### haruspex

Suppose one or more of the roots has an even denominator. What does that tell you about factors of k? Is that possible?

5. Aug 25, 2016

### Hiero

@haruspex,
I don't think this is the direction you were hinting at, but your post made me realize a way to show that the answer cannot be 4.
EDIT: There was a mistake in my logic, so I removed it.
Sorry @haruspex I haven't understood your hint.

Last edited: Aug 25, 2016
6. Aug 25, 2016

### haruspex

Yes, that's the line I was suggesting.
Sadly, it doesn't. If you could show both were even then it would.

Instead, consider that you can now rewrite the polynomial as a product of terms like (six-ti), where each ti is an integer and each si is an odd integer, and each si, ti pair is coprime.
Is this in reduced form? Would multiplying it out produce the original polynomial exactly, with no further need for cancellation?
Can any ti be even?
(This is effectively what you did next, showing the numerator must be odd. But unfortunately I didn't quote that part, and now you've deleted it I don't know if your logic was right there - I did not check it. If it was, you can proceed to the next step.)

Last edited: Aug 25, 2016
7. Aug 25, 2016

### Hiero

That was the mindless blunder which caused me to delete my reasoning a few minutes before you replied.
(My apologies, I should not have deleted it if I knew you were replying.)
I understand this.
I do not understand this. When you say "this," are you referring to the expression (s1x - t1)(s2x - t2)(s3x - t3)(s4x - t4) ? If so, then I am not sure what "reduced form" means in this context.
No, all ti must be odd in order for the constant term on the polynomial to be odd.

8. Aug 25, 2016

### haruspex

Yes, sorry, my sentence order made it confusing. I meant after multiplying it out.

Right.
Now consider the k(r1r2(r3+r4)+r3r4(r1+r2)) term. Rewrite that using the s and t integers. Can that be odd?

9. Aug 25, 2016

### Hiero

Interesting,

Using ri = ti/si and k = s1s2s3s4 I get,

k(r1r2(r3+r4)+r3r4(r1+r2)) = t1t2(s4t3+s3t4) + t3t4(s2t1+s1t2)

We know all si and all ti are odd.
Since the sum of two odd numbers is always even, the above expression is even, which contradicts what is required.

In doing this, though, haven't we assumed that all ri are rational?
So then this only shows that there can't be 4 rational roots, unless I'm mistaken?

10. Aug 25, 2016

### haruspex

What if you factor out any irrational roots?

11. Aug 25, 2016

### Hiero

We've shown there cannot be 4 rational roots.

It's obviously impossible for there to be 3 rational roots and 1 irrational root, because then kr1r2r3r4 will not be an integer.

Suppose r1 and r2 are irrational roots and r3 and r4 are rational roots. I think we can simply say that r1+r2 is irrational* therefore k(r1r2(r3+r4)+r3r4(r1+r2)) is also irrational (and thus not an odd integer). (It works the same for the case of 3 irrational roots.)

*(EDIT: The only problem I see is if r2=-r1, but then r1r2 is irrational and so kr1r2r3r4 will be irrational)

I tried it by leaving the rational roots factored out as you suggested, but things got a bit confusing. Is the above reasoning good enough?

Last edited: Aug 25, 2016
12. Aug 25, 2016

### haruspex

What about √2 and -√2 (x2-2)?

13. Aug 25, 2016

### Hiero

Oops, good point.

May I just say that this problem comes from an hour long test with 19 other problems, so you get an average of 3 minutes per problem... With that being said, is there any "easy" way to see the answer?

14. Aug 25, 2016

### haruspex

I guess there must be, but it is not obvious to me.
My next step was to consider factorisation as a quadratic with irrational roots and two linear factors expressing rational roots. It should be possible to use similar arguments as before, maybe using polynomial coefficients we've not made use of yet.

15. Aug 27, 2016

### epenguin

Although roots may be irrational, to give integral coefficients is it not the case that they cannot be just any irrational numbers? Wouldn't they have to involve rational numbers and square roots of rational numbers at most? In any case I don't think we need the nature of the roots, Rather it suffices that the expressions for coefficients that you use involve sums and products of roots that are rational.

16. Aug 28, 2016

### pasmith

There are either 0, 2 or 4 rational roots.

Suppose there are at least two. We only care about parity of coefficients, so writing O for odd and E for even the linear factors for two of these roots must take the form (Ox - O). The product of these factors is then OOx^2 - (O + O)x + OO = Ox^2 - Ex + O.

Now if the quartic is to have odd coefficients the other quadratic factor (whether factorisable over the rationals or not) must be (Ox^2 + Ux + O) where U is either odd or even. Now the x^2 coefficient of the product (Ox^2 + Ux + O)(Ox^2 - Ex + O) is (OO - UE + OO) = O + E + O = E, which is a contradiction.

17. Aug 28, 2016

### SammyS

Staff Emeritus
That's true for real roots, not true for rational roots.

Take a degree 3 polynomial having three irrational roots & multiply by some binomial having a rational root.

18. Aug 28, 2016

### SammyS

Staff Emeritus
What if you just check to see if there can at least one rational root.

Suppose there is one rational root, expressed as -t/s . Then polynomial, P can be factored as:

$P(x)=(ax^3+bx^2+cx+d)(sx+t)\,,\$ with a, b, c, d, s, and t, all being integers.​

Expand that. You can show that it's not possible for all coefficients of P to be odd integers.