How to prove $\forall a,b,x \in \mathbb N \exists c,d \in \mathbb N\setminus${$a,b$}$: (x-a)(x-b) = (x-c) (x-d)$

Question

The statement $\forall a,b,x \in \mathbb N \exists c,d \in \mathbb N\setminus${$a, b$}$: (x-a)(x-b) = (x-c) (x-d)$ seems to be true for me. But I really have no idea how to prove it formally. Is there a way?

Answer 1

This is not true; Here is a trivial counter example: $a = b = 0$. We attempt to find non-zero $c,d$ such that $(x-c)(x-d) = x^2$. By applying the distributive property twice, we find

$$x^2 - (c+d)x + cd = x^2$$ $$-(c+d)x + cd = 0$$ $$(c+d)x = cd$$

Obviously, for all $x \in \Bbb R$, no non-zero $cd$ satisfy this.

Answer 2

The statement $(x-a)(x-b)=(x-c)(x-d)$ simplifies to $x^2-(a+b)x+ab=x^2-(c+d)x+cd$ so $a+b=c+d$ and $ab=cd$. But then consider $a=1,b=2$. $2$ has no other factorization in the natural numbers, so this is false.

Worth noting that this is also false in the integers, for a similar reason. The only other factorization of $2$ in the integers is $2=(-1)(-2)$. However, this does not satisfy $a+b=c+d$.