Točno
16. listopada 2015. 16:39 (8 godine, 9 mjeseci)
Let
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
,
![b](/media/m/e/e/c/eec0d7323095a1f2101fc1a74d069df6.png)
,
![c](/media/m/e/a/3/ea344283b6fa26e4a02989dd1fb52a51.png)
,
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
,
![e](/media/m/1/6/f/16f8978af0f3c64c3eb112a539ba73dd.png)
,
![f](/media/m/9/9/8/99891073047c7d6941fc8c6a39a75cf2.png)
be positive integers and let
![S = a+b+c+d+e+f](/media/m/3/c/b/3cb573a2c95155ddd39f6859850789b6.png)
.
Suppose that the number
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
divides
![abc+def](/media/m/8/3/b/83bd572cb0a03cd2c90263caa6e36697.png)
and
![ab+bc+ca-de-ef-df](/media/m/f/1/c/f1c8076d5c429cd59205256d19483812.png)
. Prove that
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
is composite.
%V0
Let $a$, $b$, $c$, $d$, $e$, $f$ be positive integers and let $S = a+b+c+d+e+f$.
Suppose that the number $S$ divides $abc+def$ and $ab+bc+ca-de-ef-df$. Prove that $S$ is composite.
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Pretpostavimo suprotno,
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
je prost.
Neka je
![P(x) = (x+a)(x+b)(x+c), Q(x) = (x+d)(x+e)(x+f)](/media/m/1/1/9/11942169c745a3958a92ad6a24a98c13.png)
Iz pretpostavki zadatka vrijedi za svaki
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
cijeli broj:
![S \mid P(x) + Q(-x)](/media/m/5/3/3/533ea8261b906839da6a0693163522cf.png)
![x = d \implies S \mid P(d) = (d+a)(d+b)(d+c) \underset{BSOMP}{\implies} S \mid d+a](/media/m/2/4/0/2402025599dfbd6e83a274c2a6ec25e2.png)
Sto je apsurdno jer
%V0
Pretpostavimo suprotno, $S$ je prost.
Neka je $P(x) = (x+a)(x+b)(x+c), Q(x) = (x+d)(x+e)(x+f)$
Iz pretpostavki zadatka vrijedi za svaki $x$ cijeli broj: $ S \mid P(x) + Q(-x)$
$x = d \implies S \mid P(d) = (d+a)(d+b)(d+c) \underset{BSOMP}{\implies} S \mid d+a $
Sto je apsurdno jer $0 < d+a < S$
28. listopada 2015. 13:42 | grga | Točno |