$(x-a)(x-b)(x-c)(x-d)=ex$ We can verify that $x=125,162,343$ are the roots of equation $(x-105)(x-210)(x-315)=2584x$.
My question is,Could you find five positive integers $a,b,c,d,e$, which $(x-a)(x-b)(x-c)(x-d)=ex$ has four positive integer roots?
Thanks in advance.
 A: a. A parametric solution can be given to,
$$(x-a)(x-b)(x-c)(x-d) = ex$$
such that $a,b,c,d$ and all four roots $x_i$ are all positive integers as,
{$a,b,c,d$} = {$2n,\; 2n+1,\; (n+1)(n+3),\;2(n+1)(n+2)$}, and $e = n(n-1)(n+1)(n+2)(2n+3)$  
b. In general, this quartic problem can be solved by two quadratics. One must find six positive integers $a,b,c,d,p,q$ that obey,
$(a-p)(b-p)(c-pq)+p(a-p)(1-q)(c-dq)+p(1-q)(b-dq)(c-dq)=0\tag{1}$
a quadratic in $d$, such that,
$abc-pqx(a+b+c+d-p-dq-x)=0\tag{2}$
a quadratic in $x$ has positive integer roots. Then the four roots of,
$$(x-a)(x-b)(x-c)(x-d) = \frac{(p-a)(p-b)(p-c)(p-d)}{p}x\tag{3}$$
are $p,dq,x_1, x_2$ where the $x_i$ are the two roots of (2). If these four roots are integers, then the RHS of (3) obviously is also an integer.
Proof: Factor out $(x-p)$ from (3) and substituting $x = dq$ will yield (1). For the other two roots, eliminate $x$ between (2) and (3), and you will get also get (1). 
P.S. Its quite easy for Mathematica to find small $a,b,c,d,p,q$ which led me to the parametrization above. However, one must avoid trivial solutions such as,
$$(a-p)(b-p)(c-p)(d-p)(a-dq)(b-dq)(c-dq)(q-1) = 0$$
A: Thank you for your attention, some one has offered me some examples:
$(x-5)(x-6)(x-16)(x-18)=84x,x=4,9,12,20;$
$(x-3)(x-4)(x-16)(x-22)=280x,x=2,8,11,24;$
$(x-3)(x-6)(x-32)(x-44)=2520x,x=2,11,24,48.$
