Square Pair Circles
Problem
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.
Getting Started
For example
25 = 0^{2} + ( ±5)^{2} = ( ±5)^{2} + 0^{2} = ( ±3)^{2} + ( ±4)^{2} = ( ±4)^{2} + ( ±3)^{2}
and this shows that there are 12 points with integer coordinates on the circle x^{2}+ y^{2} = 5^{2} with centre at the origin and radius 5 units.
Student Solutions
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.
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 $
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 $n$.