### Be Reasonable

Prove that sqrt2, sqrt3 and sqrt5 cannot be terms of ANY arithmetic progression.

In y = ax +b when are a, -b/a, b in arithmetic progression. The polynomial y = ax^2 + bx + c has roots r1 and r2. Can a, r1, b, r2 and c be in arithmetic progression?

### Summats Clear

Find the sum, f(n), of the first n terms of the sequence: 0, 1, 1, 2, 2, 3, 3........p, p, p +1, p + 1,..... Prove that f(a + b) - f(a - b) = ab.

# Prime AP

##### Stage: 5 Challenge Level:

Show that if three prime numbers, all greater than 3, form an arithmetic progression then the common difference of the progression is divisible by 6. Find some examples of three primes which include the number 3 and form an AP, and show that in every such cases the common difference is not divisible by 6.

This solution comes from Yatir of Maccabim-Reut High-School, Israel.

Let's say that the 3 primes in the AP are:

P, Q,and S the common difference is d. So we have: P, P+d, P+2d .

We are working with primes greater than 3 so they all have to be odd and d must be even. This is because the difference between 2 odds is always even as (2n + 1) - (2k + 1) = 2(n - k) .

I'm going to work modulus 6: even number have residues of: 0, 2, 4 (mod 6) and odd numbers have residues of: 1, 3, 5 (mod 6).

Our prime numbers must be be congruent to 1 or 5 (mod 6), because if they were congruent to 3 they would be divisible by 3 and thus not prime numbers.

Lets say that is congruent to 1 (mod 6) so is congruent to either:

1 + 0 = 1 (mod 6)
1 + 2 = 3 (mod 6)
1 + 4 = 5 (mod 6).

Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 4 (mod 6).

And is congruent to either:

1 + 2 $\times$ 0 = 1 (mod 6)
1 + 2 $\times$ 4 = 1 + 7 = 9 = 3 (mod 6)

Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so must be congruent to 0 (mod 6)

Lets say that P is congruent to 5 (mod 6). So P + d is congruent to either:

5 + 0 = 5 (mod 6)
5 + 2 = 7 = 1 (mod 6)
5 + 4 = 9 = 3 (mod 6)

Because P + d is a prime number it can't be congruent to 3 (mod 6) so d must be congruent to either 0 or 2 (mod 6).

And is congruent to either:

5 + 2 $\times$ 0 = 5 (mod 6)
5 + 2 $\times$ 2 = 1 + 4 = 9 = 3 (mod 6)

Because P + 2d is a prime number as well it can't be congruent to 3 (mod 6), so d must be congruent to 0 (mod 6)

Following from all of this d must be congruent, in all cases, to 0 (mod 6), meaning it gives a remainder 0 when divided by 6. So d is divisible by 6, hence proved.

Examples from APs where one of the prime numbers is 3

3, 5, 7 (d = 2)
3, 7, 11 (d = 4)
3, 11, 19 (d = 8)
3, 13, 23 (d = 10)
3, 17, 31 (d = 14)
3, 23, 43 (d = 20)

In these examples none of the differences is divisible by 6 but is this true in general for AP's containing 3. Yes because if the first number is 3, and the common difference is divisible by 6, call this difference 6k, then the second number is 3 + 6k which is divisible by 3 so it is not a prime. Hence no AP of 3 primes exists which has common difference divisible by 6.