No questions found
Complete the list of the first six square numbers.
$1^2 = 1$ $2^2 = \text{.......}$ $3^2 = 9$ $4^2 = \text{.....}$ $5^2 = \text{.......}$ $6^2 = 36$
535
553
530
(a) Work out
(i) $9^2$, ............................................................. [1]
(ii) $40^2$. ............................................................ [1]
(b) Show that $9^2 + 40^2 = 41^2$. [2]
508
571
580
When $a^2 + b^2 = c^2$ then $(a, b, c)$ is a 3-square set. $a, b$ and $c$ are positive integers.
Example
In \textbf{Question 2(b)}, $a = 9$, $b = 40$ and $c = 41$. $9^2 + 40^2 = 41^2$, so $(9, 40, 41)$ is a 3-square set.
When $a^2 + b^2 = c^2$ then $c = \sqrt{a^2 + b^2}$.
Use this formula and any patterns you notice to complete the table on the next page for 3-square sets.
[Table_1]:
| $a$ | $b$ | $c$ |
|---|---|---|
| 3 | 4 | 5 |
| 5 | 12 | 13 |
| 7 | 24 | 25 |
| 9 | 40 | 41 |
| 11 | 60 | |
| 13 | 84 | 85 |
| | 112 | 113 |
| | 144 | |
| 19 | | 181 |
| 21 | | 221 |
| 25 | 312 | 313 |
550
563
548
When $a^2 + b^2 + c^2 = d^2$ then $(a, b, c, d)$ is a 4-square set. It is possible to make a 4-square set using two rows in the table.
Example
From the table
row two $5^2 + 12^2 = 13^2$
row six $13^2 + 84^2 = 85^2$
Replace $13^2$ in the second equation with $5^2 + 12^2$ from the first equation: $5^2 + 12^2 + 84^2 = 85^2$.
So $(5, 12, 84, 85)$ is a 4-square set.
Use the same method with rows from the table to find two more 4-square sets.
568
541
524
(a) Show that (6, 12, 12, 18) is a 4-square set.
(b) $k$ is any positive integer greater than 1.
If $(ka, kb, kc, kd)$ is a 4-square set, then $(ka)^2 + (kb)^2 + (kc)^2 = (kd)^2$.
Show that $(a, b, c, d)$ must also be a 4-square set.
(c) The numbers in the 4-square set (6, 12, 12, 18) have common factors.
(i) Find a common factor of 6, 12, 12 and 18 that is greater than 1.
(ii) Use (6, 12, 12, 18) and part (b) to find a 4-square set where $a, b, c$ and $d$ do not have a common factor greater than 1.
585
577
565
Here is another method for finding a 4-square set $(a, b, c, d)$.
Choose two positive integers $a$ and $b$ with $a$ less than $b$.
Then $c = \frac{a^2 + b^2 - 1}{2}$ and $d = \frac{a^2 + b^2 + 1}{2}$ make the 4-square set $(a, b, c, d)$.
(a) Use this to find a 4-square set when
(i) $a = 2$ and $b = 3$,
(2, 3, \text{.........} , \text{.........}) \text{ [3]}
(ii) $a = 2$ and $d = 43$.
(2, \text{.........} , \text{.........} , 43) \text{ [3]}
(b) (i) Use your answers to \textit{part (a)} and any patterns you notice to complete the table for 4-square sets that start with 2.
[Table_1]
(ii) Write down an equation connecting $c$ and $d$.
\text{............................................................} \text{ [1]}
(c) When $a$ and $b$ are both even then $c = \frac{a^2 + b^2 - 1}{2}$ and $d = \frac{a^2 + b^2 + 1}{2}$ do not give a 4-square set.
Give an example to show this. \text{ [2]}
(d) When $a$ and $b$ are both odd there are no 4-square sets.
In a 4-square set, $d = 23$.
(i) Show that $a^2 + b^2 = 45$. \text{ [1]}
(ii) Find a 4-square set when $d = 23$.
(\text{.........} , \text{.........} , \text{.........} , 23) \text{ [2]}
508
513
505
