Točno
17. kolovoza 2012. 19:35 (11 godine, 11 mjeseci)
Determine all pairs
![(x,y)](/media/m/c/9/1/c91aec4078b932368ded863349deaec5.png)
of positive integers satisfying the equation
%V0
Determine all pairs $(x,y)$ of positive integers satisfying the equation $$x!+y!=x^{y}\text{.}$$
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Promatrat cemo 5 slucajeva.
1
![x=1](/media/m/3/4/9/3491fdc1148836187540039de445a211.png)
2
![x=2](/media/m/e/f/d/efddd5fd04b20952bbeb8ccce7ca7220.png)
3
![y=1](/media/m/a/6/b/a6b934a34e7980d65e88d6a899236b07.png)
4
![y=2](/media/m/9/5/3/95318e1d58a9252bd9fe81dd9b30f35c.png)
5
![x>2](/media/m/a/a/8/aa8e411d2afe44472bdcde1edc33b9e4.png)
i
![y>2](/media/m/8/c/f/8cfd190b8d4f40998de51589797bb180.png)
1.
![y!=0](/media/m/d/7/5/d757b7eb1d9a84958ae4fae495ca43c1.png)
nema rjesenja
2.
![2 + y!=2^y](/media/m/7/9/f/79f4e945a70613c6fd945fb342be0a5a.png)
![2+y! \equiv 2^y \pmod 4](/media/m/3/e/8/3e83ab1b263e9686378f7e1adf6b1704.png)
ako
![y \geqslant 4](/media/m/3/a/8/3a83b2d98343d7f3df341d248767e2e1.png)
dobivamo
![2 \equiv 0 \pmod 4](/media/m/1/1/9/1199f6f61fc4a762999f06c1059eb45e.png)
, dakle
![y<4](/media/m/0/d/4/0d41afe6798b2fc2a38244b0ee9a9e13.png)
Provjerom dobivamo da su
![y=2](/media/m/9/5/3/95318e1d58a9252bd9fe81dd9b30f35c.png)
i
![y=3](/media/m/4/f/9/4f92e05b4ba94fe5240c3c1d619d2987.png)
rjesenja.
3.
![x! + 1=x](/media/m/d/5/2/d526b1cde0fc5dea088678352d4bde53.png)
![1 \equiv 0 \pmod x](/media/m/6/2/e/62e46955b7f390f30c362f9b238acd18.png)
dakle jedino moguce rjesenja je
![x=1](/media/m/3/4/9/3491fdc1148836187540039de445a211.png)
, a to ynamo da nema rjesenja.
4.
![x! + 2=x^2](/media/m/2/f/9/2f90e2dc63a1a04be11397174e86b014.png)
5.Iz
![x>2](/media/m/a/a/8/aa8e411d2afe44472bdcde1edc33b9e4.png)
i
![y>2](/media/m/8/c/f/8cfd190b8d4f40998de51589797bb180.png)
dobivamo
![x! \equiv 0 \pmod 6](/media/m/2/b/c/2bc58cf8f1a8ad7f1630137cf3fb1d77.png)
i
![y! \equiv 0 \pmod 6](/media/m/9/b/5/9b586500589a031ea61756d641f40fff.png)
, pa zato i
![x^y \equiv 0 \pmod 6 \Rightarrow x \equiv 0 \pmod 6](/media/m/a/a/b/aab5d24b372435fb5d814e5688877604.png)
Neka je
![(p_n)_{n \geqslant 1}](/media/m/4/a/d/4ad6ff14e897ac6aaa25165c7cb1dbd5.png)
niz prostih brojeva.
Pretpostavimo da smo dokazali
![x \equiv 0 \pmod {p_1p_2p_3...p_n}](/media/m/5/3/e/53e51b7aa5093bd4e6912de8f4a61885.png)
i
![y \geqslant 2p_n](/media/m/6/2/5/625b5e15bd7e629d96f8f1c62a0b29b8.png)
Znamo da postoji prost broj izmedi
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
i
![2n](/media/m/d/2/d/d2da874dc9bc356be9468cdbd57fbfdf.png)
pa znamo da je
![p_{n+1}<2p_n \Rightarrow y! \equiv 0 \pmod {p_{n+1}}](/media/m/e/6/6/e66978930c8f1d712d860099b0436335.png)
Takoder
![p_1p_2...p_n \geqslant 2p_n \Rightarrow x! \equiv 0 \pmod {p_{n+1}}](/media/m/3/b/0/3b03276fcbd415381f697255504d4971.png)
![x^y \equiv 0 \pmod {p_{n+1}} \newline x \equiv 0 \pmod {p_{n+1}}](/media/m/6/f/a/6fa16aa99379d72296c04092e5d294b7.png)
![p_1p_2...p_np_{n+1} \geqslant 2p_{n+1} \Rightarrow x! \equiv 0 \pmod {p_{n+1}^2}](/media/m/1/6/6/16620c75d59f9f0758973cfa5963ba14.png)
![x! + y! \equiv x^y \pmod {p_{n+1}^2} \newline y! \equiv 0 \pmod {p_{n+1}^2} \newline y \geqslant 2p_{n+1}](/media/m/5/8/2/5826fae175255e5087f55108f72f078a.png)
Dakle ovaj slucaj nema rjesenja.
Jedina rjesenja su uredeni parovi
![(2,2)](/media/m/d/4/f/d4f09538bafe4b511101ff7e041f45c7.png)
i
%V0
Promatrat cemo 5 slucajeva.
1 $x=1$
2 $x=2$
3 $y=1$
4 $y=2$
5 $x>2$ i $y>2$
1. $1 + y!=1$
$y!=0$ nema rjesenja
2. $2 + y!=2^y$
$2+y! \equiv 2^y \pmod 4 $ ako $y \geqslant 4$ dobivamo $2 \equiv 0 \pmod 4$, dakle $y<4$
Provjerom dobivamo da su $y=2$ i $y=3$ rjesenja.
3. $x! + 1=x$
$1 \equiv 0 \pmod x$ dakle jedino moguce rjesenja je $x=1$, a to ynamo da nema rjesenja.
4.$x! + 2=x^2$
$2 \equiv 0 \pmod x \Rightarrow x=2$
5.Iz $x>2$ i $y>2$ dobivamo $x! \equiv 0 \pmod 6$ i $y! \equiv 0 \pmod 6$, pa zato i $x^y \equiv 0 \pmod 6 \Rightarrow x \equiv 0 \pmod 6$
$x! + y! \equiv x^y \pmod {36} \newline y! \equiv 0 \pmod {36} \newline y \geqslant 6$
Neka je $(p_n)_{n \geqslant 1}$ niz prostih brojeva.
Pretpostavimo da smo dokazali $x \equiv 0 \pmod {p_1p_2p_3...p_n}$ i $y \geqslant 2p_n$
Znamo da postoji prost broj izmedi $n$ i $2n$ pa znamo da je $p_{n+1}<2p_n \Rightarrow y! \equiv 0 \pmod {p_{n+1}}$
Takoder $p_1p_2...p_n \geqslant 2p_n \Rightarrow x! \equiv 0 \pmod {p_{n+1}}$
$x^y \equiv 0 \pmod {p_{n+1}} \newline x \equiv 0 \pmod {p_{n+1}}$
$p_1p_2...p_np_{n+1} \geqslant 2p_{n+1} \Rightarrow x! \equiv 0 \pmod {p_{n+1}^2}$
$x! + y! \equiv x^y \pmod {p_{n+1}^2} \newline y! \equiv 0 \pmod {p_{n+1}^2} \newline y \geqslant 2p_{n+1}$
Dakle ovaj slucaj nema rjesenja.
Jedina rjesenja su uredeni parovi $(2,2)$ i $(2,3)$
16. prosinca 2012. 21:37 | grga | Točno |