Try this beautiful problem from Singapore Mathematical Olympiad, SMO, 2010 - Problem 7 based on the combination of equations.
Find the sum of all the positive integers p such that the expression (x-p) (x - 13) + 4 can be expressed in the form (x+q) (x+r) for distinct integers q and r.
Basic Algebra
Combination of Terms
Generator of a group
Answer: 26
Singapore Mathematical Olympiad
Challenges and Thrills - Pre College Mathematics
If you got stuck in this problem start this problem using this hint :
Start with the given hint
(x-p) (x-13) +4 = (x+q)(x+r)
Let's try to minimize the expression by taking x= -q
so , (-q -p)(-q -13) = -4 , which becomes (q+p) (q+13) = -4
As it is already given p and q are integers we can come up with many cases .
Try to find out the different cases we can have .............................
Starting after the last hint :
p+q = 4 and q+13 = -1 ......................................(1)
q+p = -4 and q +13 = 1.................(2)
p+q = 2 and q+13 = -2 and ..............................(3)
p+q = -2 and q+13 = 2 ..............................(4)
For 1st case its simple calculation that we get q = -14 and p = 8
The initial expression becomes (x-p) (x-13) +4 = (x-14) (x-17)
For 2 nd case :
q= -12 and p = 8
so the initial expression becomes : (x-p)(x-13)+4 = (x-9)(x-12)
try the rest of the cases............
Now let's talk about 3rd case ,
q= -15 , p = 17 and
hence (x - p) (x - 13) +4 = \(( x - 15)^2\) which is not true to this problem .
For last case , we obtain q = -11, q = 9 so the initial
(x- p) (x-13) +4 = \((x-11)^2\)
So p is 8 and 18 which add upto 8 +18 = 26 -(Answer)
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