Show A=2017!+2017+2017^2+2017^3+...+2017^2016+2017^2017 is not perfect square

[Roger.Robert]

Show that \(\displaystyle A=2017!+2017+2017^{2}+2017^{3}+2017^{4}+...+2017^{2016}+2017^{2017}\) is not a perfect square.
(from a 5th grade Olympiad)
I tried factoring 2017 but it doesn't seem to get me anywhere.

 

[Roger.Robert]

Nevermind I solved it, if I factors 2017 the second factor is bigger than 2017 so not a perfect square , idk why I didn't notice at first.

 

[Dr.Peterson]

Show that \(\displaystyle A=2017!+2017+2017^{2}+2017^{3}+2017^{4}+...+2017^{2016}+2017^{2017}\) is not a perfect square.
(from a 5th grade Olympiad)
I tried factoring 2017 but it doesn't seem to get me anywhere.

Nevermind I solved it, if I factors 2017 the second factor is bigger than 2017 so not a perfect square , idk why I didn't notice at first.

There may be more worth talking about here!

I presume you didn't really mean that a factor of 2017 is bigger than 2017 itself, which makes no sense; perhaps if you state the gist of your argument (what did you factor, and what was the final step leading to the conclusion?), it can help clarify your own thinking, and help others who are reading this.

Possibly you meant to say that you factored 2017 out of A; clearly you can do that, but how does the result allow you to say not only that A is not the square of 2017, but also that it is not the square of any integer at all?