Factorising after adding a square I have been thinking about it for quite some time but am unable to find an answer.
Let $a,b,c,d,e$ be any distinct natural numbers. Will the relation :
$(x-a)(x-b)+c^2=(x-d)(x-e)$ 
ever hold? I am unable to find any such case. We can say that
$(x-2)(x-4)+1^2=(x-3)(x-3)$ but here $d,e$ are not distinct. So, is there any example of the given form or is there any proof that it will never hold. Any kind of help will be highly appreciated.
 A: Expanding you get
$$a+b=d+e$$
$$ab+c^2=de$$
Let $$m=\frac{a+b}{2}=\frac{d+e}{2}$$ 
Then, there exists $f,g$ such that
$$a=m-f, b=m+f, d=m-g, e=m+g$$
with $m,f,g$ either integers, or half integers.
The relation then reduces to
$$g^2+c^2= f^2$$
This leads to an infinite class of solutions.
Pick $g,c,f$ a pytagorean triple and pick $m$ any integer larger than $f$ and $g$. Then
$$a=m-f, b=m+f, d=m-g, e=m+g, c$$
works.
Example If $(f,c,g)=(3,4,5)$ and $m=6$ we have
$$a=1, b=11, c=4, d=3, f=9$$
and indeed
$$(x-1)(x-11)+16=(x-3)(x-9)$$
P.S. These represent all solutions with $m,f,g$ integers. The solutions with $m,f,g$ half integers lead to the equation
$$(2g)^2+(2c)^2= (2f)^2$$
where $2g, 2f$ are odd integers and $2c$ is even integer. Then, if you pick $m=\frac{2k+1}{2}$ you have the solution
$$a=m-f, b=m+f, d=m-g, e=m+g, c$$
A: I meant not much different decision. Using Pythagorean triples we get a solution depending on two parameters. Although the system itself:
$$\left\{\begin{aligned}&a+b=d+e\\&ab+c^2=de\end{aligned}\right.$$
Can have solutions depend on 3 parameters, for example:
$$a=s(p-k)$$
$$b=p^2+s(p+s-k)$$
$$d=s(p+s-k)$$
$$e=p^2+s(p-k)$$
$$c=ps$$
$p,s,k$ - integers, any sign.
