Thread:

1. polynomial equation

calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(P(x)-x)^2$

2. Originally Posted by jacks
calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(p(x)-x)^2$
If you put x=0 in this equation then you get $(p(-1))^2+1=0$. This cannot happen if the polynomial has real coefficients. So there are no such real polynomials. If complex coefficients are allowed, then $p(-1)=\pm\sqrt{-1}$, but I don't see where to go from there.

3. Originally Posted by jacks
calculate all polynomial $p(x)$ for which $(p(x))^2+(p(x-1))^2+1=(P(x)-x)^2$
If we presume that p(x) is going to have at least real coefficients we can expland the definig equation to get
$\displaystyle p(x+1) = \frac{1}{2}(x + 1) - \frac{p^2(x)}{2(x + 1)}$

The above says that p(x) must contain a factor of x + 1 and since no real p(0) exists (ala Opalg) we must also require that p(x) contains a factor of x as well, so we have
$\displaystyle p(x) = x(x + 1)q(x)$

where q(x) is a polynomial. This expression needs to be run through the original equation to get restrictions on q(x), but I can't make any sense out of the resulting equation.

-Dan
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9488778114318848, "perplexity_flag": "head"}