Rational roots of 4th degree polynomial with odd coefficents

[Hiero]

Homework Statement

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?

Homework Equations

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

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.

 

[Charles Link]

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.

 

[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.

 

[haruspex]

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

 

[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.

 

[haruspex]

If the product r1r2r3r4 (in reduced form) has an even denominator, then k would need to be even for k(r1r2r3r4) to be an integer (contradicting the first quoted equation).

Yes, that's the line I was suggesting.

r1r2r3r4 must have an odd numerator and an odd denominator, but this contradicts the assumption that r1r2r3r4 is in reduced form.

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 (s_ix-t_i), where each t_i is an integer and each s_i is an odd integer, and each s_i, t_i 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 t_i 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.)

 

[Hiero]

Sadly, it doesn't. If you could show both were even then it would.

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.)

Instead, consider that you can now rewrite the polynomial as a product of terms like (s_ix-t_i), where each t_i is an integer and each s_i is an odd integer, and each s_i, t_i pair is coprime.

I understand this.

Is this in reduced form?

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.

Can any t_i be even?

No, all t_i must be odd in order for the constant term on the polynomial to be odd.

 

[haruspex]

When you say "this,"

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

No, all ti must be odd in order for the constant term on the polynomial to be odd.

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

 

[Hiero]

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

Interesting,

Using r_i = t_i/s_i and k = s₁s₂s₃s₄ I get,

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

We know all s_i and all t_i 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 r_i are rational?
So then this only shows that there can't be 4 rational roots, unless I'm mistaken?

 

[haruspex]

So then this only shows that there can't be 4 rational roots,

What if you factor out any irrational roots?

 

[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?

 

[haruspex]

The only problem I see is if r2=-r1, but then r1r2 is irrational

What about √2 and -√2 (x²-2)?

 

[Hiero]

What about √2 and -√2 (x²-2)?

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?

 

[haruspex]

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?

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.

 

[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.

 

[pasmith]

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.

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.

 

[SammyS]

There are either 0, 2 or 4 rational roots.
...

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.

 

[SammyS]

Interesting,

Using r_i = t_i/s_i and k = s₁s₂s₃s₄ I get,

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

We know all s_i and all t_i 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 r_i are rational?
So then this only shows that there can't be 4 rational roots, unless I'm mistaken?

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.