**1. How many of the three digit numbers that can be made from all of the the digits 1, 3 and 5 (used only once each) are prime?**

None of them! All permutations of 135 are multiples of 3.

**2. 12345 can be expressed as the sum of two primes in exactly one way. What is the larger of the two primes whose sum is 12345?**

12343. Two odds add up to an even, so one of the primes must be 2.

**3a. Find all prime numbers p such that 2p+1 is a perfect square.**

Answer: there are none. Here's why: Assume 2p+1 is a perfect square. 2p+1 is odd, and all odd perfect squares are equivalent to 1 (mod 4), so 2p is equivalent to 0 (mod 4), in other words it is a multiple of 4. That means p is even, so it must be 2, the only even prime number. But 2*2+1=5, which is not a perfect square.

**3b. Find all prime numbers p such that 2p+1 is a perfect cube.**

Answer: p=13 is the only solution. Here's why (solution by Chris Braun):

Assume 2p+1 is a cube.

2p = q^{3}-1 = (q-1)(q^{2}+q+1)

Both factors of the RHS are at least 2, because q is odd and bigger than 1. 2p is the product of two primes, so if 2p=ab, and both a and b are larger than 1, then a=2 and b=p or vice versa. q-1 is even, and so it must be 2, which makes q^{2}+q+1 equal to 13.

Alternative solution: A quick test reveals p can't be 2, so p is odd.

Let p=2k+1.

2p+1 = 2(2k+1)+1 = 4k+3 = q^{3} for some integer, q.

q^{3} = 3 (mod 4), so q must also be equivalent to 3 (mod 4).

(To see this, observe that 0^{3}, 1^{3}, and 2^{3} have residues (mod

4) of 0, 1, and 0, respectively, not 3 as required.)

let q = 4n+3. Then,

2p+1 = 4k+3 = q^{3} = (4n+3)^{3} = 64n^{3}+144n^{2}+108n+27.

Subtract 1 from both sides, and divide by 2 to solve for p in terms of n:

p=32n^{3}+72n^{2}+54n+13, which factors nicely as

p=(2n+1)(16n^{2}+28n+13), so if p is prime, n must equal zero (or -1, but a quick test eliminates that), so p=13 is the only prime number that satisfies 2p+1=x^{3}

**4a. Let n be an integer greater than 6. Prove that if n-1 and n+1 are both prime, then n²(n²+16) is divisible by 720.**

n is divisible by 6, so n^{2} is divisible by 36 and (n^{2}+16) is divisible by 4, so n^{2}(n^{2}+16) is divisible by 144. Then 5 is the only additional factor we need to find in order to show the result.

If n=1 (mod 5) then n-1 isn't prime, and if n=-1 (mod 5) then n+1 isn't prime, so n must be 0, 2, or 3 (mod 5), so we'll consider those three cases:

case n=0 (mod 5): n^{2} is divisible by 5, and the result follows.

case n=2 (mod 5): n^{2}+16 is 2^{2}+1=0 (mod 5), and the result follows.

case n=3 (mod 5): n^{2}+16 is 3^{2}+1=0 (mod 5), and the result follows.

**4b. Is the converse true? That is, if n**^{2}**(n**^{2}**+16) is divisible by 720, then are n-1 and n+1 both prime?**

No, the converse is not true, because n^{2}(n^{2}+16) is divisible by 720 whenever n is a multiple of 6 and n===0, 2, or 3 (mod 5).

To find a counterexample, n-1 or n+1 must be the product of prime factors larger than 5, so n=48 is the smallest counterexample.

If n=48, then n^{2}(n^{2}+16) is divisible by 720, but n+1=49 is not prime.

**5. Let a be the integer whose base 10 representation consists of 119 ones. Prove that a is not prime.**

Let's look at some smaller examples of numbers consisting of a composite number of ones. 111111 has 6 1's, so it is the product of 010101 and 11. 111111111111111 has 15 1's, so it is the product of 001001001001001 and 111.

119 = 7 * 17, so it's the product of 00000010000001...0000001 and 1111111.

In general, if k is a factor of n, then (x^{k-1}+x^{k-2}+...+1) is a factor of (x^{n-1}+x^{n-2}+...+1)

**6. Prove composite: n**^{4}+4^{n}, n>1.

If n is even, then n^{4}+4^{n} is even, so consider the case in which n is odd.

Let n=2k+1, so now we need to show that n^{4}+4^{2k+1} is composite for any k > 0.

n^{4}+4^{2k+1} = (n^{2}+2^{2k+1}-2^{k+1}n)(n^{2}+2^{2k+1}+2^{k+1}n)

This clever factorization uses the trick that:

a^{4}+4b^{4} = (a^{2}+2b^{2}-2ab)(a^{2}+2b^{2}+2ab)

and since n^{4}+4^{2k+1} = n^{4}+(4)(2^{4k}), we can let a=n and b=2^{k} to achieve the desired result.