Točno
12. rujna 2016. 21:41 (7 godine, 10 mjeseci)
Dva prosta broja, $p,q(p>q)$ , zadovoljavaju $p^q + 9q^6 = k^2$ gdje je $k$ prirodan broj. Pronađi $p + q$.
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![p^q = k^2 - 9q^6 = (k-3q^3)(k+3q^3)](/media/m/9/5/5/9551d816e1d2cb4632503b135580b88a.png)
Prema tome, imamo za
![a+b=q](/media/m/f/9/8/f984e64f307935573770aa135b2c0a7f.png)
:
![p^a = k - 3q^3](/media/m/f/6/6/f6652d9f6c7d18178e90d54231d9b987.png)
![p^b = k + 3q^3](/media/m/e/0/7/e0781d4d6fbf22ec5981a846f25b067f.png)
![p^a + p^b = 2k](/media/m/6/b/8/6b8aaadb7901ec16978a3085612d8220.png)
odakle ili
![p \ | \ 2k](/media/m/7/a/5/7a545bb41435a8f477fa798d59f35b55.png)
ili
![a = 0](/media/m/8/9/f/89ff6f70cad2e8926501b9a1eae13028.png)
. Očito,
![p = 2](/media/m/3/9/7/397dde586af3c79a7b51514f9fd8cf8b.png)
nije rješenje zbog
![p > q](/media/m/3/f/1/3f12495e2e525546fe93124bbd32e893.png)
pa imamo
![p \geq 3](/media/m/9/f/0/9f0c935d7738faea35703b7f65c29563.png)
Nadalje, ako
![p \ | \ k](/media/m/4/8/a/48a4730ef1e5a6956f45bea0ac7d40e8.png)
tada
![k = np](/media/m/e/2/4/e24850d5c174d6510967ce0a1420ed14.png)
pa
![p^q = n^2p^2 - 9q^6](/media/m/3/e/6/3e6639a7bf03c9d10649bba0da247af6.png)
pa
![p \ | \ 9q^6](/media/m/c/9/b/c9bc5e4fb2042a816d8eed04ec9f0990.png)
.
No, tada bi bilo ili
![p = q](/media/m/4/b/1/4b1af15bb4c2a4ad643dd13456d6b4fb.png)
što nije moguće po uvjetu
![p > q](/media/m/3/f/1/3f12495e2e525546fe93124bbd32e893.png)
ili
![p=3 \implies q = 2](/media/m/e/9/b/e9b6be21d03d5253cf75fa5f2e4b1cf1.png)
što nije rješenje (a time imamo i
![p \geq 5](/media/m/f/3/e/f3e75cfd3e2db080098f3fa1c97e064c.png)
)
Zaključujemo da
![a =0](/media/m/1/3/f/13ff359b1273558b7358218885417855.png)
.
Tada:
![k = 3q^3 + 1](/media/m/8/9/4/89475187f6e449e64ff971695b1589b0.png)
pa
![p^q = p^b = 6q^3 + 1](/media/m/a/c/d/acdbd0c1ee08e3d3dbf2fde3caa27992.png)
![p^q = 6q^3+1 \geq 5^q \implies q < 4 \implies q \in \{2, 3\}](/media/m/f/4/3/f431a9d2754154e6dc4c463cd0e24371.png)
Ako
![q = 2](/media/m/9/3/b/93b12d1f901f77fbc376f1e9a5d6f22e.png)
tada
![p^2 = 49](/media/m/f/7/3/f737060b44a883f7dd4826264b14b529.png)
pa
![(p, q) = (7, 2)](/media/m/6/6/9/669c963ddfa87fcf1ee4bb4838661a73.png)
Ako
![q = 3](/media/m/a/4/6/a46d972f1d33b832a1a69edc29cb4f91.png)
tada
![p^3 = 163](/media/m/3/e/a/3ea10734a50207a06bf83bac68e21eac.png)
pa nema rješenja.
Prema tome
![p+q = 9](/media/m/9/f/7/9f7b9867ab3a89e1adb9b20affa19540.png)
.
%V0
$$p^q = k^2 - 9q^6 = (k-3q^3)(k+3q^3)$$
Prema tome, imamo za $a+b=q$:
$$p^a = k - 3q^3$$
$$p^b = k + 3q^3$$
$$p^a + p^b = 2k$$
odakle ili $p \ | \ 2k$ ili $a = 0$. Očito, $p = 2$ nije rješenje zbog $p > q$ pa imamo $p \geq 3$
Nadalje, ako $p \ | \ k$ tada $k = np$ pa $p^q = n^2p^2 - 9q^6$ pa $p \ | \ 9q^6$.
No, tada bi bilo ili $p = q$ što nije moguće po uvjetu $ p > q$ ili $p=3 \implies q = 2$ što nije rješenje (a time imamo i $p \geq 5$)
Zaključujemo da $a =0$.
Tada: $k = 3q^3 + 1$ pa $p^q = p^b = 6q^3 + 1$
$$p^q = 6q^3+1 \geq 5^q \implies q < 4 \implies q \in \{2, 3\}$$
Ako $q = 2$ tada $p^2 = 49$ pa $(p, q) = (7, 2)$
Ako $q = 3$ tada $p^3 = 163$ pa nema rješenja.
Prema tome $p+q = 9$.