Proving that a+b+c+d is a Multiple of 4 Given (x-a)(x-b)(x-c)(x-d)-4 = 0

[b0mb0nika]

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]

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.

 

[b0mb0nika]

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]

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.

 

[mathwonk]

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.

 

[matt grime]

remembering the distinct nature of a,b,c,d was what held me up cos i couldn't think of any compelling reason to know more about r-a etc other than they were divisors of 4. But thinking about how I might get a counter example led me to say, well, what if r-a=1, r-b=1, r-c=2, r-d=2, then we get a contradiction: a+b+c+d=4r-6. That's how the extra hypothesis came to my attention.

And the idea to think of divisors of 4 came about from helping my students find a reason why x^5+x-1 has no rational roots (any such root is an integer, and a divisor of 1, hence +/-1, and neither of those is actually a root). If that weren't in the back of my mind I don't know how long it would have taken me.