High School Mathematics Extensions/Primes/Problem Set/Solutions

Question 1
Is there a rule to determine whether a 3-digit number is divisible by 11? If yes, derive that rule.

 Solution 

Let x be a 3-digit number We have
 * $$x = 100a + 10b + c \!$$

now
 * $$x \equiv a + 10b + c \equiv a - b + c \pmod{11} \!$$

We can conclude a 3-digit number is divisible by 11 if and only if the sum of first and last digit minus the second is divisible by 11.

Question 2
Show that p, p + 2 and p + 4 cannot all be primes. (p a positive integer and is great than 3)

Solution 

We look at the arithmetic mod 3, then p slotted into one of three categories
 * 1st category
 * $$p \equiv 0 \pmod{3} \!$$
 * we deduce p is not prime, as it's a multiple of 3
 * 2nd category
 * $$p \equiv 1 \pmod{3} \!$$
 * $$p + 2\equiv 0 \pmod{3} \!$$
 * so p + 2 is not prime
 * 3rd category
 * $$p \equiv 2 \pmod{3} \!$$
 * $$p + 4\equiv 0 \pmod{3} \!$$
 * therefore p + 4 is not prime

Therefore p, p + 2 and p + 4 cannot all be primes.

Question 3
Find x

\begin{matrix} x \equiv 1^7 + 2^7 + 3^7 + 4^7 + 5^7 + 6^7 + 7^7 \ \pmod{7}\\ \end{matrix} $$

 Solution 

Notice that
 * $$-a \equiv 7-a \pmod 7 \!$$.

Then
 * $$1^7 \equiv (7-6)^7 \equiv (-6)^7 \equiv -(6^7) \pmod 7 \!$$.

Likewise,
 * $$2^7 \equiv -5^7 \pmod 7 \!$$

and
 * $$3^7 \equiv -4^7 \pmod 7 \!$$.

Then


 * $$x \!$$
 * $$\equiv 1^7 + 2^7 + 3^7 + 4^7 + 5^7 + 6^7 + 7^7 \!$$
 * $$\equiv 1^7 + 2^7 + 3^7 - 3^7 - 2^7 -1^7 + 7^7 \!$$
 * $$\equiv 0 \pmod{7} \!$$
 * }
 * $$\equiv 1^7 + 2^7 + 3^7 - 3^7 - 2^7 -1^7 + 7^7 \!$$
 * $$\equiv 0 \pmod{7} \!$$
 * }
 * $$\equiv 0 \pmod{7} \!$$
 * }
 * }

Question 4
9. Show that there are no integers x and y such that
 * $$x^2 - 5y^2 = 3 \!$$

Solution 

Look at the equation mod 5, we have
 * $$x^2 = 3 \pmod{ 5} \!$$

but

therefore there does not exist a x such that
 * $$1^2 \equiv 1 \!$$
 * $$2^2 \equiv 4 \!$$
 * $$3^2 \equiv 4 \!$$
 * $$4^2 \equiv 1 \!$$
 * }
 * $$4^2 \equiv 1 \!$$
 * }
 * }
 * $$x^2 \equiv 3 \pmod{5} \!$$

Question 5
Let p be a prime number. Show that

(a)

(p-1)! \equiv -1\ \pmod{p} $$ where

n! = 1 \cdot 2 \cdot 3 \cdots (n-1) \cdot n $$ E.g. 3! = 1&times;2&times;3 = 6

(b) Hence, show that
 * $$\sqrt{-1} \equiv \frac{p - 1}{2}! \pmod{p}$$

for p &equiv; 1 (mod 4)

 Solution 

a) If p = 2, then it's obvious. So we suppose p is an odd prime. Since p is prime, some deep thought will reveal that every distinct element multiplied by some other element will give 1. Since
 * $$(p - 1)! = (p - 1)(p - 2)(p - 3) \cdots 2 \!$$

we can pair up the inverses (two numbers that multiply to give one), and (p - 1) has itself as an inverse, therefore it's the only element not "eliminated"

as required.
 * $$(p - 1)! \equiv (p - 1) \equiv - 1 \!$$
 * }

b) From part a)
 * $$-1 \equiv (p - 1)! \!$$

since p = 4k + 1 for some positive integer k, (p - 1)! has 4k terms
 * $$-1 = 1\times2 \times 3 \times \cdots 2k \times (-2k) \cdots \times(- 3) \times (- 2) \times (- 1) $$

there are an even number of minuses on the right hand side, so
 * $$-1 = (1\times2 \times 3 \times \cdots 2k)^2$$

it follows
 * $$\sqrt{-1} = 1\times 2\times 3\times ... 2k$$

and finally we note that p = 4k + 1, we can conclude
 * $$\sqrt{-1} = \frac{p - 1}{2}!$$