High School Mathematics Extensions/Primes/Problem Set/Solutions

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100 percents.svg Primes
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At the moment, the main focus is on authoring the main content of each chapter. Therefore this exercise solutions section may be out of date and appear disorganised.

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Question 1[edit]

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[edit]

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[edit]

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}  \!

Question 4[edit]

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

1^2 \equiv 1 \!
2^2 \equiv 4 \!
3^2 \equiv 4 \!
4^2 \equiv 1 \!

therefore there does not exist a x such that

x^2 \equiv 3 \pmod{5} \!

Question 5[edit]

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! = 1×2×3 = 6

(b) Hence, show that

\sqrt{-1} \equiv \frac{p - 1}{2}! \pmod{p}

for p ≡ 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"

(p - 1)! \equiv (p - 1) \equiv - 1 \!

as required.

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}!