Neocijenjeno
6. studenoga 2023. 22:37 (8 mjeseci, 1 tjedan)
Odredi sve parove
prirodnih brojeva za koje vrijedi
Odredi sve parove $(a, b)$ prirodnih brojeva za koje vrijedi
\[ a^5 + a^4 = 7^b - 1 \text. \]
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![(a^2+a+1)(a^3-a+1)=7^b](/media/m/a/8/0/a80017612aa3797b788e45073a025e8d.png)
![M(a^3-a+1,a^2+a+1)=M(-a^2-2a+1,a^2+a+1)=d](/media/m/f/0/a/f0af3bc235379a148c5d0f3169bb9440.png)
![d|a^2+a+1+a^2+2a-1, d|2a^2+3a](/media/m/5/c/5/5c52b8f8b90d21457d9bd1c814a90c6d.png)
![d|2a^2+3a-2a^2-2a-2,d|a-2](/media/m/8/6/1/861f5e54bc3effd1140634b446e7570e.png)
![d|2a^2+3a-2a(a-2),d|7a](/media/m/b/2/e/b2e7a78e2791f5c823a3564bc5dbbd10.png)
![d|7a-7(a-2),d|14, d=1,2,7,14](/media/m/e/5/1/e51c7e83466ea1f1991efdc42c9ef5e7.png)
zbog parnosti oba dviju strana, ![d=1,7](/media/m/3/3/e/33e0d7de627aa70e66358c6630b1259e.png)
![d=1,a= \pm 1, 7^b=3, \Rightarrow a \ge 2](/media/m/5/2/f/52fe25e3674ca4cc6d665c4f3db23c8f.png)
![d=7](/media/m/f/2/4/f249eafb90cd564b8aba7d0aa726402b.png)
![a^3-a+1>a^2+a+1 \Rightarrow a^2+a+1=7](/media/m/8/5/4/854a71cad3cc972c98521604be4b2564.png)
![a(a+1)=6, a=2,b=2](/media/m/d/f/1/df106b93a3a399a6c993e2129365afd8.png)
Jedino rješenje je
$(a^2+a+1)(a^3-a+1)=7^b$
$M(a^3-a+1,a^2+a+1)=M(-a^2-2a+1,a^2+a+1)=d$
$d|a^2+a+1+a^2+2a-1, d|2a^2+3a$
$d|2a^2+3a-2a^2-2a-2,d|a-2$
$d|2a^2+3a-2a(a-2),d|7a$
$d|7a-7(a-2),d|14, d=1,2,7,14$
zbog parnosti oba dviju strana, $d=1,7$
$d=1,a= \pm 1, 7^b=3, \Rightarrow a \ge 2$
$d=7$
$a^3-a+1>a^2+a+1 \Rightarrow a^2+a+1=7$
$a(a+1)=6, a=2,b=2$
Jedino rješenje je $(a,b)=(2,2)$