Exercise (3.5).7

(i) Factorise 3397301 (you don’t need to factorise this into prime factors).

(ii) Solve $x^2 + 164x− 3 397 301 = 0$ without using the quadratic formula.

(i) $\lceil \sqrt{3397301} \rceil = 1844$, and $1844^2 - 3397301 = 3035$ which is not a perfect square.

$1845^2 - 3397301 = 82^2$, and so $3397301 = (1845-82)(1845+82) = 1927 \times 1763$.

The factorisation $3397301 = 1927 \times 1763$ is sufficient.

(ii) We notice that $1927-1763=164$, we have

$$ \begin{align} 0 & = x^2 + 164x - 3397301 \\ \\ & = x^2 +(1927-1763)x - (1927  \times 1763) \\ \\ &= (x+1927)(x-1763) \end{align} $$

And so the solutions are $x=-1927$ and $x+1763$.

Exercise (3.5).6

Factorise $18 861 649$. Hence or otherwise solve the quadratic

$$ x^2 − 18 861 649 = 0$$

We find that $\sqrt{18861649} = 4343$ exactly. So we can work on 4343.

$\lceil \sqrt{4343} \rceil = 66$, and $66^2 - 4343 = 13$ which is not a perfect square.

By additional trials we find $72^2 - 4343 = 29^2$, and so $4343 = (72-29)(72+29) = 43 \times 101$.

So the factorisation is $18861649 = 43^2 \times 101^2$.

Using the above factorisation we have 

$$ \begin{align} 0 & = x^2 - 18861649 \\ \\  &=  x^2 - (43 \times 101)^2 \\ \\ & = (x - 4343)(x+4343) \end{align} $$

And so $x = \pm 4343$.

Exercise (3.5).5

 (i) Factorise each of the following integers:

(a) 713 (b) 1271 (c) 403

(ii) Solve the quadratic equation

$$ 403x^2 + 1271x + 713 = 0 $$

leaving your answer in surd form.

(iii) Simplify the following fractions:

$$ \frac{713}{1271}, \quad \frac{403}{1271}, \quad \frac{403}{713} $$

(i) (a) We have $\lceil \sqrt{713} \rceil = 27$, and $27^2 - 713 = 4^2$, and so $713 = (27-4)(27+4) = 23 \times 31$.

The factorisation is $713 = 23 \times 31$.

(b)  $\lceil \sqrt{1271} \rceil = 36$, and $36^2 - 1271 = 5^2$, and so $1271 = (36-5)(36+5) = 31 \times 41$.

The factorisation is $1271 = 31 \times 41$

(c)  $\lceil \sqrt{403} \rceil = 21$, and $21^2 - 403 = 38$, which is not a perfect square.

$22^2 - 403 = 9^2$, and so $403 = (22-9)(22+9) = 13 \times 31$.

The factorisation is $403 = 13 \times 31$

(ii) We can use the factorisations above

$$ \begin{align} 0 & = 403x^2 + 1271x + 713 \\ \\ & = (13 \cdot 31)x^2 + (31 \cdot 41)x + (23 \cdot 31) \\ \\  & = 13x^2 +41x +23  \end{align} $$

Using the general solution $x= \frac{-b \pm \sqrt{b^2-4ac}}{2a}$, we have

$$ \begin{align} x & = \frac{-41 \pm \sqrt{41^2-4(13)(23)}}{2(41)} \\ \\  x & =  \frac{-41 \pm \sqrt{485}}{26}  \end{align} $$

(iii) $$ \frac{713}{1271} = \frac{23 \cdot 31}{31 \cdot 41} = \frac{23}{41}$$

$$ \frac{403}{1271} = \frac{13 \cdot 31}{31 \cdot 41} = \frac{13}{41} $$

$$ \frac{403}{713} = \frac{13 \cdot 31}{23 \cdot 31} = \frac{13}{23} $$

Exercise (3.5).4

Factorize 1 236 519.

$\lceil \sqrt{1236519} \rceil = 1112$, and $1112^2 - 1236519 = 5^2$, and so $1236519 = 1112^2 0 5^2 = (1112+5)(1112-5) = 1117 \times 1107$.

We notice that the digits of 1107 sum to 9, and so is divisible by 9. In fact, we find it is divisible by 27, leaving prime 41. We also find the 1117 is prime, by testing up to $\sqrt{1117}$.

So the factorisation is $1236519 = 3^3 \times 41 \times 1117$.

Exercise (3.5).3

Let $n$ be an odd integer. Show that

$$ \left( \frac{n+1}{2} \right )^2 - \left( \frac{n-1}{2} \right )^2 = n $$

$$ \begin{align} \left( \frac{n+1}{2} \right )^2 - \left( \frac{n-1}{2} \right )^2 & = \left( \frac{n^2 + 2n +1}{4} \right ) - \left( \frac{n^2 -2n +1}{4} \right ) \\ \\ & = \left( \frac{4n}{4} \right ) \\ \\ & = n \end{align} $$

Exercise (3.5).2

Factorise the following into their prime factors:

(a) 9271 (b) 2146 (c) 2 974 791

(a) $\lceil \sqrt{9271} \rceil = 97$, and $ 97^2 - 9271 = 138 $, which is not a perfect square.

$ 98^2 - 9271 = 333 $, which is not a perfect square.

$ 99^2 - 9271 = 530 $, which is not a perfect square.

$ 100^2 - 9271 = 27^2 $, and so $9271 = 100^2 - 27^2 = (100-27)(100+27) = 73 \times 127$.

The prime factorisation is $9271 = 73 \times 127$.

(b)  Since 2146 is even, we can factor out 2, and work on the odd 1073.

$\lceil \sqrt{1073} \rceil = 33$, and $ 33^2 - 1073 = 4^2 $, and so $1073 = 33^2 - 4^2 = (33+4)(33-4) = 37 \times 29$.

The prime factorisation is $2146 = 2 \times 29 \times 37$.

(c) The Fermat factorisation approach doesn't yield easy results for several trials and so we reduce the number by factoring out 3, and working on 991597 as a smaller odd number.

$\lceil \sqrt{991597} \rceil = 996$, and $996^2 - 991597 = 419$ which is not a perfect square.

We keep trying until we find $1001^2 - 991597 = 102^2$, and so $991597 = 1001^2 - 102^2 = (1001 - 102)(1001 + 102) = 1103 \times 899 = 1103 \times 29 \times 31 $.

The prime factorisation is $2974791 = 3 \times 29 \times 31 \times 1103$.

Exercise (3.5).1

Factorise the following integers into their prime factors:

(a) 299 (b) 851 (c) 10403 (d) 2479

We will use Fermat difference of two squares factorisation method.

(a)  $\lceil \sqrt(299) \rceil = 18$, and $18^2-299 = 5^2$.

So $299 = 18^2 - 5^2 = (18+5)(18-5) = 23 \times 13$.

So the prime factorisation is $299=23 \times 13$.

(b) $\lceil \sqrt(851) \rceil = 30$, and $30^2-851 = 7^2$.

So $851 = 30^2 - 7^2 = (30+7)(30-7) = 37 \times 23$.

So the prime factorisation is $851 = 37 \times 23$.

(c) $\lceil \sqrt(10403) \rceil = 102$, and $102^2-10403 = 1^2$.

So $10403 = (102^2+1)(102^2-1) = 103 \times 101$.

So the prime factorisation is $10403 = 103 \times 101$.

(d) $\lceil \sqrt(2479) \rceil = 50$, and $50^2-2479 = 21$ which is not a perfect square.

So we try $51^2-2479=122$, also not a perfect square.

So we try $52^2 - 2479 = 15^2$, and so $2479 = 52^2 - 15^2 = (52+15)(52-15)= 67 \times 37$.

So $2479 = 67 \times 37$.