Točno
1. travnja 2016. 20:51 (8 godine, 3 mjeseci)
Find all ordered pairs
![(a,b)](/media/m/e/2/6/e263229694cdbeb908488db2d0351f0a.png)
of positive integers for which the numbers
![\dfrac{a^3b-1}{a+1}](/media/m/6/1/7/6176135fd940484d49602888ee243945.png)
and
![\dfrac{b^3a+1}{b-1}](/media/m/f/9/5/f957a55bbbaaa7a8a978fe3176c491b1.png)
are both positive integers.
%V0
Find all ordered pairs $(a,b)$ of positive integers for which the numbers $\dfrac{a^3b-1}{a+1}$ and $\dfrac{b^3a+1}{b-1}$ are both positive integers.
Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
![\dfrac{b^3a+1}{b-1} = ab^2 + ab + a + \dfrac{a+1}{b-1} \implies b-1 | a+1 \implies b-1 | b +1](/media/m/c/5/f/c5f4cb71ca59f1859e6e02f6956407e1.png)
![b-1](/media/m/c/5/a/c5a6116902e1f56d4ca4716c62834ccd.png)
je djelitelj od
![b+1](/media/m/7/1/1/711e1b74fd6da02f784dc73d62cf6df0.png)
pa, pošto
![b-1 \neq b+1](/media/m/a/a/1/aa19a75f8aa144f64b7cf7e69eb8f7df.png)
, slijedi
![b-1 \leq \dfrac{b+1}{2}](/media/m/e/4/5/e4594e7874028ed1f6084e1ea6aa7283.png)
ili sređivanjem
![b \leq 3](/media/m/b/4/2/b4252b4cce7d0f4b89d6cd12bdb02958.png)
pa promatramo mogućnosti
1)
![b=3](/media/m/c/8/3/c83c34bf47581eaefc5874ca5949581c.png)
U tom slučaju
![b-1|b+1 \iff 2|4](/media/m/3/f/2/3f289de51733eb52347782798ab993c9.png)
što vrijedi
također
![b-1\mid a+1\mid b+1 \iff 2 \mid a+1 \mid 4](/media/m/1/b/2/1b286afaa8f8d0286f39e78db8fd68e6.png)
pa gledamo mogućnosti za
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
1.1)
![a + 1 = 2 \implies a = 1](/media/m/8/7/3/873414b0e30d812e1dd9875b784c48ad.png)
pa provjerom imamo uređen par
![(a,b) = (1, 3)](/media/m/c/2/a/c2a1f99b09e5835768258cda272bfb37.png)
kao jedno rješenje
1.2)
![a + 2= 4 \implies a = 3](/media/m/e/a/0/ea041a2d9e026632aee9765c3a64e94e.png)
pa provjerom imamo
![(a,b) = (3,3)](/media/m/4/c/1/4c1a36851cb5db220c5ba9c28bcee203.png)
kao još jedno rješenje
2)
![b=2](/media/m/f/8/d/f8d26bb35153bc6de8fe79d6cfb9b6fd.png)
U tom slučaju
![b-1|b+1 \iff 1|3](/media/m/6/d/4/6d4cb03b71cea8b7d7d2898bdd5e220c.png)
što vrijedi
također
![b-1\mid a+1\mid b+1 \iff 1 \mid a+1 \mid 3](/media/m/d/c/4/dc4b20eafc4f86096b0cb989674fe25b.png)
pa gledamo mogućnosti za
![a](/media/m/6/d/2/6d2832265560bb67cf117009608524f6.png)
2.1)
![a + 1 = 1 \implies a = 0](/media/m/3/f/3/3f3a979a7ba6af5084a9d78f06496199.png)
što je nemoguće jer
![0 \notin N](/media/m/3/b/0/3b064a6f42f1a527a4a62e432e8318ef.png)
2.2)
![a + 1 = 3 \implies a = 2](/media/m/f/4/3/f43eaaf5fd1350216f3c764765612cc4.png)
pa provjerom imamo
![(a,b) = (2,2)](/media/m/5/0/6/506fc817170d55ecfb47acba537228da.png)
kao rješenje
3)
![b=1](/media/m/b/c/f/bcf77481be15a83071b6665a535b39b0.png)
što je nemoguće jer tada u
![\dfrac{b^3a+1}{b-1}](/media/m/f/9/5/f957a55bbbaaa7a8a978fe3176c491b1.png)
imamo
![0](/media/m/7/b/8/7b8b0b058cf5852d38ded7a42d6292f5.png)
kao nazivnik
Sva moguća rješenja su uređeni parovi
%V0
$ \dfrac{a^3b-1}{a+1} = a^2b - ab + b - \dfrac{b+1}{a+1} \implies a+1 | b+1$
$ \dfrac{b^3a+1}{b-1} = ab^2 + ab + a + \dfrac{a+1}{b-1} \implies b-1 | a+1 \implies b-1 | b +1$
$b-1$ je djelitelj od $b+1$ pa, pošto $b-1 \neq b+1$, slijedi $b-1 \leq \dfrac{b+1}{2}$ ili sređivanjem $b \leq 3$ pa promatramo mogućnosti
1) $b=3$
U tom slučaju $b-1|b+1 \iff 2|4$ što vrijedi
također $b-1\mid a+1\mid b+1 \iff 2 \mid a+1 \mid 4$ pa gledamo mogućnosti za $a$
1.1) $a + 1 = 2 \implies a = 1$ pa provjerom imamo uređen par $(a,b) = (1, 3)$ kao jedno rješenje
1.2) $a + 2= 4 \implies a = 3$ pa provjerom imamo $(a,b) = (3,3)$ kao još jedno rješenje
2) $b=2$
U tom slučaju $b-1|b+1 \iff 1|3$ što vrijedi
također $b-1\mid a+1\mid b+1 \iff 1 \mid a+1 \mid 3$ pa gledamo mogućnosti za $a$
2.1) $a + 1 = 1 \implies a = 0$ što je nemoguće jer $0 \notin N$
2.2) $a + 1 = 3 \implies a = 2$ pa provjerom imamo $(a,b) = (2,2)$ kao rješenje
3) $b=1$ što je nemoguće jer tada u $\dfrac{b^3a+1}{b-1}$ imamo $0$ kao nazivnik
Sva moguća rješenja su uređeni parovi $(a,b) = (1,3), (2,2), (3,3)$