root problem.

Posts: 791
Joined: Aug 4, 2011

by hemangsarkar, Sep 7, 2012, 6:50 AM

Q) let $a,b,cd$ be distinct integers, such that
$(x-a)(x-b)(x-c)(x-d) = 4$ has an integer root $r$ , prove that -

$a + b + c + d = 4r$ .

solution :
assume that $a < b < c < d$

so, $-d < -c < -b < -a$

and $r-d < r-c < r-b < r-a$

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

and since $(r-a), (r-b), (r-c), (r-d)$ are all distinct integers,

we "must" have $(r-a) = 2$

$(r-b) = 1$

$(r-c) = -1$

$(r-d) = -2$

on adding we get $4r = a + b + c + d$