Question 1158926
<pre>

The factorization of x² + x - n must be in the form

(x - p)(x + q) where p and q are positive integers.

We must have this identity for all values of x:

 x² + x - n = (x - p)(x + q)
 x² + x - n = x² + qx - px - pq
 x² + x - n = x² + (q - p)x - pq

So q - p = 1
      pq = n

       q = p + 1

     p(p+1) = n

 {{{1<=n<=1000}}}

{{{1<=p(p+1)<=1000}}}

{{{1<=p^2+p<=1000}}}

{{{matrix(3,5,

1<=p^2+p,"",and,"",p^2+p<=1000,
p^2+p>=1,"",and,"",p^2+p-1000<=0,
p^2+p-1>=0,"","","","")}}}       

The only positive zero of the polynomial on the left is approximately 0.6,
and the p²+p-1 is positive for and only for positive values of integer p greater
than or equal to 1.

The only negative zero of the polynomial on the right is approximately 31.1,
and the p²+p-1000 is negative only for positive values of p less than or equal
to 31.  So p<=31

Therefore there are 31 ways the polynomial x² + x - n can be factored as the
product of two linear factors with integer coefficients.

     factorization   when    and 
     (x-p)(x+q)        p=     n=
--------------------------------
 1.  (x-1)(x+2)         1     2
 2.  (x-2)(x+3)         2     6
 3.  (x-3)(x+4)         3    12 
…
29.  (x-29)(x+30)      29   870
30.  (x-30)(x+31)      30   930
31.  (x-31)(x+32)      31   992

Edwin</pre>