# Homework Help: Which primes p satisfy p^2|5^p^2+1?

1. Nov 13, 2009

### DianaSagita

1. The problem statement, all variables and given/known data

First of all, hi everyone!
I'm quite new in number theory, and need help on this one badly...

Determine all prime numbers p so p2 divides 5p2+1.
2. Relevant equations

Euler's theorem: If a and m are coprimes then $$a^{\varphi(m)}\equiv 1 (mod\ m)$$
where $$\varphi(m)$$ (Euler's function) denotes number of positive integers which are coprime with m and not greater than given int m.

Special: Fermat's little theorem... if p is prime, p and a coprimes, then $$a^{p-1}\equiv 1 (mod\ p)$$

...and... $$\varphi(p^{2})=p^{2}-p$$

3. The attempt at a solution

Know one solution p=3, but I got it by assumption. :((

2. Nov 14, 2009

### icystrike

Hi there (=
I'm also new to number theory.
May i check with you if the primes that satisfy the above condition is 2 and 3?
I am writing my full proof as i am living in(asia) different GMT from yours.

<Question : Determine all prime numbers p so p2 divides 5p2+1.>

By Fermat's Little Theorem,

$$5^p$$ is congruent to 5 (mod p) (1)

Which suggest that $$5^{p.p} = 5^{p}^{2}$$ is congruent to 5 (mod p) as stated in (1).

Since $$5^{p}^{2}+1$$is divisible to $$p^{2}$$ ;

Therefore, $$5^{p}^{2}+1$$ is divisible to p ;

5+1 (mod p) is congruent to 0 (mod p)

Thus, 6 is a multiple of p .

pk = 6=2*3 with k is an element of integer.

By Euclid's Lemma,

Therefore , p can be either 2 or 3.

Last edited: Nov 14, 2009
3. Nov 14, 2009

### DianaSagita

Well, thanks a lot, it seems correct!
Meanwhile, I did it too...
According to Euler's Theorem:
$$5^{p^{2}-p}\equiv 1(mod\ p^{2})$$
$$5^{p^{2}}\equiv 5^{p}(mod\ p^{2})$$
$$5^{p^{2}}-5^{p}\equiv 0(mod\ p^{2})$$
$$5^{p^{2}}+1-(5^{p}+1)\equiv 0(mod\ p^{2})$$
So, if $$5^{p^{2}}+1$$ is divisible by p2, then $$5^{p}+1$$ must be, too.
Now, using Little Fermat's Theorem:
$$5^{p-1}\equiv 1(mod\ p)$$
$$5^{p}\equiv 5(mod\ p)$$
$$5^{p}-5\equiv 0(mod\ p)$$
$$5^{p}+1-6\equiv 0(mod\ p)$$
If $$5^{p}+1$$ is divisible by p, then 6 must be divisible by p. The only prime numbers which satisfy this are, as you proved, p=2 and p=3. But, if we check it with 1st statement, we'll see that p=3 is unique solution in N field.

Thank you once again, cheers!

4. Nov 15, 2009

### Mentallic

For p=2,
$$p^2|5^p^2+1$$
is 156.5

Now, I haven't studied number theory, but isn't the definition of "divides" that the solution be a whole number? Which, as you can see for p=2 it is not.

5. Nov 16, 2009

### icystrike

yes you are right i've forgotten to check with the first statement (=