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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

${\displaystyle x=100a+10b+c\!}$

now

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

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

${\displaystyle p\equiv 0{\pmod {3}}\!}$

we deduce p is not prime, as it's a multiple of 3

2nd category

${\displaystyle p\equiv 1{\pmod {3}}\!}$

${\displaystyle p+2\equiv 0{\pmod {3}}\!}$

so p + 2 is not prime

3rd category

${\displaystyle p\equiv 2{\pmod {3}}\!}$

${\displaystyle p+4\equiv 0{\pmod {3}}\!}$

therefore p + 4 is not prime

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

Question 3

Find x

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

Solution

Notice that

${\displaystyle -a\equiv 7-a{\pmod {7}}\!}$.

Then

${\displaystyle 1^{7}\equiv (7-6)^{7}\equiv (-6)^{7}\equiv -(6^{7}){\pmod {7}}\!}$.

Likewise,

${\displaystyle 2^{7}\equiv -5^{7}{\pmod {7}}\!}$

and

${\displaystyle 3^{7}\equiv -4^{7}{\pmod {7}}\!}$.

Then

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

Question 4

9. Show that there are no integers x and y such that

${\displaystyle x^{2}-5y^{2}=3\!}$

Solution

Look at the equation mod 5, we have

${\displaystyle x^{2}=3{\pmod {5}}\!}$

but

${\displaystyle 1^{2}\equiv 1\!}$

${\displaystyle 2^{2}\equiv 4\!}$

${\displaystyle 3^{2}\equiv 4\!}$

${\displaystyle 4^{2}\equiv 1\!}$

therefore there does not exist a x such that

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

Question 5

Let p be a prime number. Show that

(a)

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

where

${\displaystyle n!=1\cdot 2\cdot 3\cdots (n-1)\cdot n}$

E.g. 3! = 1×2×3 = 6

(b) Hence, show that

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

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

${\displaystyle (p-1)!=(p-1)(p-2)(p-3)\cdots 2\!}$

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"

${\displaystyle (p-1)!\equiv (p-1)\equiv -1\!}$

as required.

b) From part a)

${\displaystyle -1\equiv (p-1)!\!}$

since p = 4k + 1 for some positive integer k, (p - 1)! has 4k terms

${\displaystyle -1=1\times 2\times 3\times \cdots 2k\times (-2k)\cdots \times (-3)\times (-2)\times (-1)}$

there are an even number of minuses on the right hand side, so

${\displaystyle -1=(1\times 2\times 3\times \cdots 2k)^{2}}$

it follows

${\displaystyle {\sqrt {-1}}=1\times 2\times 3\times ...2k}$

and finally we note that p = 4k + 1, we can conclude

${\displaystyle {\sqrt {-1}}={\frac {p-1}{2}}!}$