# Thread:

1. ## Prime Numbers

Here is my problem:

Let a,b be nonzero integers. Prove that (a,b) = 1 if and only if (a+b, ab) = 1.

Any suggestion on where to start?

2. You could try playing around with Bezout's identity.

(From right to left) Suppose that $\gcd(a+b,ab)=1$. There exist integers $x,y$ such that $(a+b)x+(ab)y=1$. Distributing the terms, $ax+bx+aby=1$, and $a(x+by)+bx=1$. This shows that $\gcd(a,b)$ must divide 1. Therefore, $\gcd(a,b)=1$.

You can try a similar argument for the other direction. (You may need to insert some terms into the equation by adding and subtracting the same thing.)

3. Alternatively:

$<br /> \gcd(a+b,ab) = 1 \Leftrightarrow \gcd(a+b,a) = \gcd(a+b,b) = 1 \Leftrightarrow \gcd(a,b)=1<br />$
{"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": 8, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "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": 0.895769476890564, "perplexity_flag": "head"}