Timeline for What makes this polynomial a square number?
Current License: CC BY-SA 4.0
13 events
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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 |
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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 |
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Jan 12, 2021 at 5:59 | history | answered | Ankoganit | CC BY-SA 4.0 |