Točno
27. kolovoza 2014. 17:43 (9 godine, 10 mjeseci)
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Neka je
pozitivan djeljitelj broja
,
njihov broj i
traženi oblik broja
.
Poznato je da
![\prod_{d \mid n}d = n ^{\frac{\tau(n)}{2}}](/media/m/3/b/5/3b576a3d43a103394a2b6703c6d5c7f5.png)
Dokaz:
Promatramo slučajeve
i ![2\nmid \tau(n)](/media/m/6/5/b/65bc4f33b0a8bfd437a7528bf290b078.png)
1.slučaj
Svaki djeljitelj broja
je oblika
gdje je
.
Dakle ima sveukupno
djeljitelja.
Zbog
sve djeljitelje možemo grupirati u dvočlane skupove
a kako je umnožak
i takvih skupova je
zaključujemo ![\prod_{d \mid n}d = n ^{\frac{\tau(n)}{2}}](/media/m/f/9/2/f9259913c0449f04a5300bee567b9b8d.png)
2.slučaj
Analogno
ali zbog
zaključujemo da je
potpuni kvadrat i stoga postoji
tako da je
. Sve djeljitelje osim toga grupiramo u istih
dvočlanih skupova. Na kraju je ![\prod_{d \mid n}d = n ^ {\frac{\tau(n)-1}{2}} \cdot \sqrt{n} = n ^{\frac{\tau(n)}{2}}](/media/m/a/9/0/a90fab3f230d9cc3b9216691d7741212.png)
Iz uvjeta zadatka onda redom slijedi
![n ^{\frac{\tau(n)}{2}}=n^3](/media/m/5/3/2/5327052569a07907d5f49b0a0e88b45c.png)
![\frac{\tau(n)}{2}=3](/media/m/6/8/2/682fe0e09424a08e663f00dd9f6b98ab.png)
![\tau(n)= 6](/media/m/9/4/8/9480d02327408163e5684156745531c7.png)
Također znamo da je
ali ![\tau(n)= 6= 1\cdot 6 = 2 \cdot 3](/media/m/3/2/4/324db4e254732162270e1118480aa0db.png)
Zaključujemo da je ili
ili
.
U prvom slučaju imamo
![(\alpha_1 +1 )= 6 \Rightarrow \alpha_1=5 \Rightarrow \boxed{n= p_1^5}](/media/m/8/7/0/8705a3f6b003d84ce676575fef7fbf6d.png)
a u drugom na isti način
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![\tau(n)](/media/m/9/3/5/93573657f80f39012c212de4b60f0b1f.png)
![n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}](/media/m/f/3/7/f3765738a6872d71a66a9e737daeb03b.png)
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
Poznato je da
![\prod_{d \mid n}d = n ^{\frac{\tau(n)}{2}}](/media/m/3/b/5/3b576a3d43a103394a2b6703c6d5c7f5.png)
Dokaz:
Promatramo slučajeve
![2\mid \tau(n)](/media/m/e/1/2/e12f59a634524936e56a4ac8b0a42e59.png)
![2\nmid \tau(n)](/media/m/6/5/b/65bc4f33b0a8bfd437a7528bf290b078.png)
1.slučaj
Svaki djeljitelj broja
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}](/media/m/2/a/6/2a631e959e352ef723dd4995262facbd.png)
![0 \leq \beta_i \leq \alpha_i](/media/m/a/1/0/a10dd43de5b217465bbe34bd03f76797.png)
Dakle ima sveukupno
![(\alpha_1+1)(\alpha_2 +1)\cdots (\alpha_k +1) = \tau(n)](/media/m/b/8/3/b8302c75daf952c47a54329763e073e3.png)
Zbog
![2 \mid \tau(n)](/media/m/b/1/f/b1f5d38a5b81049f12fff71b9c1ea227.png)
![(d, \frac{n}{d} )](/media/m/9/6/0/96059db0efd10fda2e1236aebb91ceba.png)
![d \cdot \frac{n}{d} = n](/media/m/e/0/8/e0823da26548c2308191d35d92a07de6.png)
![\frac{\tau(n)}{2}](/media/m/6/e/7/6e71feeca891d9800c5912753af85247.png)
![\prod_{d \mid n}d = n ^{\frac{\tau(n)}{2}}](/media/m/f/9/2/f9259913c0449f04a5300bee567b9b8d.png)
2.slučaj
Analogno
![\tau(n) = (\alpha_1+1)(\alpha_2 +1)\cdots (\alpha_k +1)](/media/m/2/0/b/20b64404d82a295ba066a6a177c6a81b.png)
![2 \nmid \tau(n)](/media/m/e/f/7/ef7364748e2d7584c0bcb1c4b13eac5f.png)
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![d](/media/m/f/7/d/f7d3dcc684965febe6006946a72e0cd3.png)
![d^2 = n](/media/m/c/9/a/c9a37cb71945fdf438d54c79fb4927b1.png)
![\frac{\tau(n)-1}{2}](/media/m/d/0/9/d098f459e009b7dacb129107e2469f02.png)
![\prod_{d \mid n}d = n ^ {\frac{\tau(n)-1}{2}} \cdot \sqrt{n} = n ^{\frac{\tau(n)}{2}}](/media/m/a/9/0/a90fab3f230d9cc3b9216691d7741212.png)
Iz uvjeta zadatka onda redom slijedi
![n ^{\frac{\tau(n)}{2}}=n^3](/media/m/5/3/2/5327052569a07907d5f49b0a0e88b45c.png)
![\frac{\tau(n)}{2}=3](/media/m/6/8/2/682fe0e09424a08e663f00dd9f6b98ab.png)
![\tau(n)= 6](/media/m/9/4/8/9480d02327408163e5684156745531c7.png)
Također znamo da je
![\tau(n) = (\alpha_1 +1)(\alpha_2 +1) \cdots (\alpha_k +1)](/media/m/9/4/4/94417ddc63d0d8bb7853bc9c92be3919.png)
![\tau(n)= 6= 1\cdot 6 = 2 \cdot 3](/media/m/3/2/4/324db4e254732162270e1118480aa0db.png)
Zaključujemo da je ili
![k=1](/media/m/f/7/1/f71077af98878d94ee3faacc57dd14b5.png)
![k=2](/media/m/b/2/c/b2c85402c049e3e48f39c947b9411e8a.png)
U prvom slučaju imamo
![(\alpha_1 +1 )= 6 \Rightarrow \alpha_1=5 \Rightarrow \boxed{n= p_1^5}](/media/m/8/7/0/8705a3f6b003d84ce676575fef7fbf6d.png)
a u drugom na isti način
![\boxed{n=p_1p_2^2}](/media/m/7/2/b/72b30290bf386864b6d797c5896f90bc.png)
Ocjene: (1)
Komentari:
ikicic, 27. kolovoza 2014. 13:52