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Jul 27, 2021 at 5:51 comment added Ankoganit @HemantAgarwal The only integer between those two is $2x^2+x+1$, so the square root has to be equal to $2x^2+x+1$. Plus, its square us supposed to be $4x^4+4x^3+4x^2+4x+4$, hence the equation.
Jul 26, 2021 at 23:18 comment added Hemant Agarwal So, I understood that the numbers we are looking for will be greater than $(2x^2+x)$ and less than $(2x^2+x+2)$ . But, how do we know that they compulsorily have to solve the equation : $(2x^2+x+1)^2=4x^4+4x^3+4x^2+4x+4$ and any number that does not satisfy this equation, cannot be the number we are looking for.
Jan 12, 2021 at 23:29 comment added Bubbler @BmyGuest Done.
Jan 12, 2021 at 23:17 vote accept Bubbler
Jan 12, 2021 at 16:31 comment added BmyGuest @Bubbler Could you add the refence solution (as community wiki maybe)?
Jan 12, 2021 at 13:37 comment added Ankoganit @pajonk when I tried to estimate the square root of the original thing with trinonials, I ran into fractions: x^2+x/2+1 and the like. The only point of scaling up by 4 was to get rid of the fractions.
Jan 12, 2021 at 13:23 comment added pajonk @Ankoganit How did you come up with the idea to multiply the polynomial by 4?
Jan 12, 2021 at 13:11 comment added Ankoganit Edited to get rid of unnecessary casework and add some more explanation.
Jan 12, 2021 at 13:10 history edited Ankoganit CC BY-SA 4.0
deleted 49 characters in body
Jan 12, 2021 at 6:11 comment added Bubbler The answer is correct, and this proof (even with unnecessary case split) seems more elegant than the reference solution!
Jan 12, 2021 at 6:10 comment added Ankoganit In retrospect, there was no need to look at positives and negatives separately: the key argument works nicely for general non-zero x. Oh well.
Jan 12, 2021 at 6:06 history edited Ankoganit CC BY-SA 4.0
edited body
Jan 12, 2021 at 5:59 history answered Ankoganit CC BY-SA 4.0