[QUOTE=bur;586538]
[I]Let p and p^2+8 be prime. Prove that p^3+4 is also prime. [/I] <snip> So only if p=3 we can have both p and p^2+8 be prime. <snip> Or is it really just about showing that it only holds for p=3 and since 3^3+4=31 is prime, that's the proof?[/QUOTE] :tu: Yup. 
Another problem came up. The exercises deal with showing that a^x is congruent to 1 (or a) mod c, where c is a composite number:
a^21 == a (mod 15) for all a This is not a problem yet, but what I'm struggling with is if the composite has the square of a prime, such as: a^6 == 1 (mod 168) for all a where gcd(a,42)=1. 168 = 3*7*8 so it has to be shown that a^6 == 1 (mod 3), (mod 7), (mod 8). Mod 3 and 7 is no problem, but what about mod 8? 8 is not prime, so Fermat's little theorem cannot be applied. I noticed that apparently [TEX]a^{p^{n1}(p1)} \equiv 1 \pmod {p^{n}}[/TEX], but have no proof for it and I doubt they want that as solution as it wasn't discussed prior. I could prove it by the fact that all odd numbers squared are == 1 (mod 8), so (a^3)^2 is as well, but that only works for mod 8. Also, a later question is: a^4 = 1 (mod 60) for all a where gcd(a,30)=1. So (mod 4=2^2) comes into it. 
The square of every odd number is congruent to 1 (mod 8).

You could simply check a^2 (mod 8) for all four possible cases of a (mod 8). Then a^6 = (a^2)^3.
Note that knowing a^2 (mod 8) means you know a^2 (mod 4). 
[QUOTE=bur;593477]
This is not a problem yet, but what I'm struggling with is if the composite has the square of a prime, such as: a^6 == 1 (mod 168) for all a where gcd(a,42)=1. 168 = 3*7*8 so it has to be shown that a^6 == 1 (mod 3), (mod 7), (mod 8). Mod 3 and 7 is no problem, but what about mod 8? 8 is not prime, so Fermat's little theorem cannot be applied. I noticed that apparently [TEX]a^{p^{n1}(p1)} \equiv 1 \pmod {p^{n}}[/TEX], but have no proof for it and I doubt they want that as solution as it wasn't discussed prior.[/QUOTE] That is [url]https://en.wikipedia.org/wiki/Euler%27s_theorem[/url] for m=p^n. When you want the best exponent for all gcd(a,n)=1 it is also a known problem (n=168 in your case). Let lambda(n)=lcm(,eulerphi(p[i]^e[i],) for n=prod(p[i]^e[i]) but if p=2 and e>2 then you can gain an additional factor of 2 on that place writing eulerphi(2^e)/2. And this is the best possible/smallest exponent what you can write for that a^lambda(n)==1 mod n for every gcd(a,n)=1. For example in your case n=2^3*3*7 and lambda(n)=lcm(2,2,6)=6 ps. maybe in maths you're using the unmodified lcm() when p=2 and e>2, the ugly case here is due to the fact that there is no primitive root mod 2^e if e>2.The proof of the best exponent is easy: just use that there is a primitive root mod m iff m=1,2,4,p^e or 2*p^e when p is an odd prime. 
[QUOTE=Nick;593483]The square of every odd number is congruent to 1 (mod 8).[/QUOTE]
I know, I wrote just that... ;) [QUOTE=Dr Sardonicus;593485]You could simply check a^2 (mod 8) for all four possible cases of a (mod 8). Then a^6 = (a^2)^3. Note that knowing a^2 (mod 8) means you know a^2 (mod 4).[/QUOTE]I didn't think of that. But in that case it still only works for powers of 2 as factors. [QUOTE]That is [URL]https://en.wikipedia.org/wiki/Euler%27s_theorem[/URL] for m=p^n[/QUOTE]Thanks, this is the answer to my question, but seeing it, I think it wasn't what the creators of the exercise had in mind since eulerphi is one chapter later. So I assume they were after answers specific to that problem i.e. divisibility of of squares by 8 and 4. Thanks for your help everyone. 
And another one:
Show that [TEX]{p1\choose k}\equiv 1^k \pmod p[/TEX] It probably shouldn't use Wilson's theorem since that only comes in the next section in the book. I tried to write it as a fraction: [TEX]{p1\choose k}=\frac{(pk)\cdot...\cdot(p1)}{k!}=\frac{(k+1)\cdot...\cdot(p1)}{(p1k)!}[/TEX] All those factors cannot be reduced (mod p) since they smaller than p. And [tex]p1\equiv1\pmod p[/tex]. I'd be thankful for a hint in the right direction. 
[QUOTE=bur;593946]I tried to write it as a fraction:
[TEX]{p1\choose k}=\frac{(pk)\cdot...\cdot(p1)}{k!}=\frac{(k+1)\cdot...\cdot(p1)}{(p1k)!}[/TEX] All those factors cannot be reduced (mod p) since they smaller than p. And [tex]p1\equiv1\pmod p[/tex].[/QUOTE] It depends what you mean by "reduced". Try subtracting p from all the factors of the numerator, so p1 turns into 1, p2 into 2 etc, and see what that gives you. 
[QUOTE=bur;593946]And another one:
Show that [TEX]{p1\choose k}\equiv 1^k \pmod p[/TEX] It probably shouldn't use Wilson's theorem since that only comes in the next section in the book.[/QUOTE] I see where you could use Wilson theorem in the problem, though those solutions would be a little ugly. But really don't see what is in the next chapter and what you could/should use. Freely use whatever theorems are in your head. Solutions without Wilson: About this problem (I know tex but not writing Latex here to save time): the precise problem: binomial(p1,k)==(1)^k mod p for 0<=k<p. the start was good, say if the answer is x then ((p1)*(p2)*...*(pk))/k!==x mod p multiple by k!, you need: (p1)*(p2)*...*(pk)==x*k! mod p, but in the left side: pi==(i) mod p for every i integer, so (1)*(2)*...*(k)==x*k! mod p k!*(1)^k==x*k! mod p, we can divide by k!, since k! is not divisible by p, so k! is relative prime to p. (1)^k==x mod p, what was needed. There is a one line solution here: (p1)*(p2)*...*(pk)/k!==(1)*(2)*...*(k)/k!==(1)^k*k!/k!==(1)^k mod p [you need somewhat more knowledge to understand this]. 
[QUOTE=bur;593946]And another one:
Show that [TEX]{p1\choose k}\equiv 1^k \pmod p[/TEX] <snip>[/QUOTE]Clearly binomial(p1,0) = 1 = (1)^0, so it's true for k = 0. Simplify the ratio binomial(p1,k+1)/binomial(p1,k) [you wind up with one linear term in the numerator and one in the denominator]. Check that the numerator and denominator are nonzero for k = 0 to p2, and are equal and opposite (mod p). 
[QUOTE=Dr Sardonicus;594023]Clearly binomial(p1,0) = 1 = (1)^0, so it's true for k = 0. Simplify the ratio binomial(p1,k+1)/binomial(p1,k) [you wind up with one linear term in the numerator and one in the denominator]. Check that the numerator and denominator are nonzero for k = 0 to p2, and are equal and opposite (mod p).[/QUOTE]
Modified your idea, close to a pure combinatorial proof: The k=0 case is trivial, so assume that 0<k<p, it is known: binomial(p1,k1)+binomial(p1,k)=binomial(p,k) [for here you don't need that p is prime] furthermore: k*binomial(p,k)=p*binomial(p1,k1) is also true, here the right side is divisible by p, so the left side also, but 0<k<p, hence binomial(p,k) is divisible by p. From this binomial(p1,k)==binomial(p1,k1)==...==(1)^k*binomial(p1,0)==(1)^k mod p, what we needed. 
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