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.
On
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.
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 $$a=m-f, b=m+f, d=m-g, e=m+g, c$$