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Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2


All powers of 5 can be written as the sum of two square numbers in different ways. Investigate the number of points with integer coordinates on circles with centres at the origin for which the square of the radius is a power of 5. For the two circles shown in red in the diagram the radius is a power of 5 and for the circles shown in black this is not so but the square of the radius is a power of 5.

Thanks to Rusi Kolev for the following solution.

To find all the points which lie on the circle k(0;r) and O(0;0) we have to use the circle equation $ x^2 + y^2 = r^2 $

If we find such real numbers x and y then the point with coordinates (x,y) lies on the cirlce.

It is given that $ r^2 =5^n$, n is a positive integer and $ n \geq 1 \rightarrow r = \surd(5^n);$ Then we have the equation$ x^2 + y^2 = 5^n \rightarrow x^2 = 5^n - y^2 \rightarrow x = \surd(5^n -y^2)$ and x and y are positive integers (belong to N).

To find all the x and y which satisfy the equation we have to try all the y's up to $ \surd5^n$.

As $x^2$ and $y^2$ are positive numbers then x and y are positive or negative. Then we can find all the positive x and y and make up combinations. For example if $ x_1 \geq 0$ and $y_1 \geq 0 $ then all the solutions are: $(x_1,y_1), (x_1,-y_1), (-x_1,y_1), (-x_1,-y_1).$.

Here are all the positive solutions for n up to 12:

n

x

y

0
1
0
0
0
1
1
2
1
1
1
2
2
5
0
2
4
3
2
3
4
2
0
5
3
11
2
3
10
5
3
5
10
3
2
11
4
25
0
4
24
7
4
20
15
4
15
20
4
7
24
4
0
25
5
55
10
5
50
25
5
41
38
5
38
41
5
25
50
5
10
55
6
125
0
6
120
35
6
117
44
6
100
75
6
75
100
6
44
117
6
35
120
6
0
125
7
278
29
7
275
50
7
250
125
7
205
190
7
190
205
7
125
250
7
50
275
7
29
278
8
625
0
8
600
175
8
585
220
8
527
336
8
500
375
8
375
500
8
336
527
8
220
585
8
175
600
8
0
625
9
1390
145
9
1375
250
9
1250
625
9
1199
718
9
1025
950
9
950
1025
9
718
1199
9
625
1250
9
250
1375
9
145
1390
10
3125
0
10
3116
237
10
3000
875
10
2925
1100
10
2635
1680
10
2500
1875
10
1875
2500
10
1680
2635
10
1100
2925
10
875
3000
10
237
3116
10
0
3125
11
6950
725
11
6875
1250
11
6469
2642
11
6250
3125
11
5995
3590
11
5125
4750
11
4750
5125
11
3590
5995
11
3125
6250
11
2642
6469
11
1250
6875
11
725
6950
12
15625
1185
12
15580
4375
12
15000
5500
12
14625
8400
12
13175
9375
12
12500
10296
12
11753
11753
12
10296
12500
12
9375
13175
12
8400
14625
12
5500
15000
12
4375
15580
12
1185
15625



What we notice is when n = 0 or 1 we have only one solution (by one solution I mean one pair of POSITIVE integers that satisfy the equation.

We can always make all the solutions by using the combinations explained above! .

When n = 2 or 3 we have 2 solutions ... so we know that for a given integer n the number of solutions is:1.

If n is even The number of positive solutions is "n + 2".

If n is odd The number of positive solutions is "n + 1".

But why is this so? Why do the solutions always increase with 1?

Let`s take n = 2k (even):

n=2

We notice the solutions:$0 \times5^1$; $5^1$ and $3 \times5^0$; $4 \times5^0$; and $4 \times5^0$; $3 \times5^0$ and $5^1$; $0 \times5^1$

n=4

$0 \times5^2$; $5^2$ and $3 \times5^1$; $4 \times5^1$; and $7 \times5^0$; $24 \times5^0$ and $24 \times5^0$; $7 \times5^0$ and $4 \times5^1$; $3 \times5^1$ and $5^2$; $0 \times5^2$

n=6

$0 \times5^2$; $5^2$ and $3 \times5^1$; $4 \times5^1$; and $7 \times5^0$; $24 \times5^0$ and $24 \times5^0$; $7 \times5^0$ and $4 \times5^1$; $3 \times5^1$ and $5^2$; $0 \times5^2$

Every next even n has all the previous solutions but raised to the +1 power plus TWO NEW solutions.

Now let`s take n = 2k + 1(odd):

n=3, (k=1).

We notice the solutions:$2 \times5^0$; $11 \times5^0$ and $5 \times5^0$; $10 \times5^0$ and $11 \times5^0$; $2 \times5^0$

n=5 (k=2)

$2 \times5^1$; $11 \times5^1$ and $5 \times5^1$; $10 \times5^1$ and $38 \times5^0$; $41 \times5^0$ and $41 \times5^0$; $38 \times5^0$ and $10 \times5^1$; $5 \times5^1$ and $11 \times5^1$; $2 \times5^1$

n=7 (k=3)

$2 \times5^2$; $11 \times5^2$ and $5 \times5^2$; $10 \times5^2$ and $38 \times5^1$; $41 \times5^1$ and $29 \times5^0$; $278 \times5^0$ and $278 \times5^0$; $29 \times5^0$; and $41 \times5^1$; $38 \times5^1$ and $10 \times5^2$; $5 \times5^2$ and $11 \times5^2$; $2 \times 5^2$

Every next odd n has all the previous solutions but raised to the +1 power plus TWO NEW solutions.

As n approaches infinity the solutions will become more and more ...

Until now we have only discussed the positive integers. If we are to make all the solutions we will use the example above: if $x_1 \geq 0$ and $y_1 \geq 0 $

then all the solutions are: $( x_1, y_1), (x_1, -y_1), (-x_1, y_1), (-x_1, -y_1).$

So if we have one solution we can make four (positive and negative) so we have to multiply by 4. Taking the formulae above:

If $n$ is even, the number of solutions "$n + 2$" becomes "$4(n + 2)$"
If $n$ is odd, the number of solutions "$n + 1$" becomes "$4(n + 1)$"

But these formulae are also not correct because if we have root 0 then there isn`t +0 and -0. We have root 0 only when $n=2k$; $k=1$,$2$,$3$....
We have $4n + 8$ solutions if we count +0 and -0. But becausethere is no difference between +0 and -0 we deduct 2 solutions for every zero root,
Solutions are: $4n + 8 - 4= 4n + 4 = 4(n + 1)$. And that`s the number of solutions for an odd $n$.