# Rational root

suppose a,b,c,d are distinct integers such that (x-a)(x-b)(x-c)(x-d)-4 =0

has a rational root r. Prove that a+b+c+d is a multiple of 4.

I've been trying to solve this for a couple of hours, but except the fact that r is an integer, b/c we have a monic polynomial I didn't get too far.

when we expand we get -(a+b+c+d) x^3 as one of the terms but i'm not sure what to do with it.

matt grime
Homework Helper
x=r is a root, and as you correctly note it is an integer. Hence

(r-a)(r-b)(r-c)(r-d)=4

Thus r-a etc are all divisors of 4, and are all distinct. The only way to have that 4 is the product of 4 distinct integers is if they are 1,-2,2,-2, Hence (r-a)+(r-b)+(r-x)+(r-c) = 0 = 4r -a-b-c-d, or a+b+c+d=4r, where r is the integer root.

This is a nice question. I've not seen one like it before.

u mean 1, -1, 2, -2.. right ?
ah this wasn't that hard... why didn't i come up with this sol thanks:)

matt grime
Homework Helper
yes, 1,-1,2,-2 slip of the fingers. It's amazing how many questions are easy when you know tha answer. They key in maths is to decide when the effort required in applying a line of reasoning without success "outweighs" its chances of being the correct answer. I just happened to be in a frame of mine where it seemed obvious to consider the factrors of 4. Took me a few seconds to figure out the exact details.

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mathwonk
Homework Helper
2020 Award
it took me more than few seconds, but my ultimate solution was the same as Matt's.

The test of whether something is "hard" is not whether it looks short when presented by someone who understands it, but whether one can think of it.

So to me this was a little harder than many such questions on here, simply because I did not get it right away.

The additional hardness for me, was that it did not follow instantly from the usual suspects, i.e. the rational root theorem.

One had to write down the correct equation, [that product equals 4], and look at it long enough to realize that prime factorization implies a strong restriction on it.

So the thing to ask yourself is how could you have thought of it yourself. I.e. ask yourself where you missed seeing it. just what little path you stopped short of going down.

Oh yes, and for me, one slipup was that I did not at first use all the hypotheses, i.e. the distinctness of the a,b,c,d. Even after I started factoring the expression (n-a)(n-b)(n-c)(n-d) = 4, I did not at first remember this fact.

So remember to ask yourself if there is an assumption you have not used. That may give you an idea of how to use it.

I also took a hint from Hurkyl, in that I did not look at Matt's post until I had it myself, as I knew he would have solved it completely and I would be denied the chance to discover the trick.

Actually discovering a solution opens a little channel in your mind in a beneficial way that reading one does not always do.

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matt grime