# How many pairs of positive integer a, b

1. Oct 9, 2004

### omicron

How many pairs of positive integer a, b are such that $$a^2 + b^2 = 121?$$

Last edited: Oct 9, 2004
2. Oct 9, 2004

### Tide

Are there any positive integers for which

$$(11-b)(11+b)$$

are perfect squares?

3. Oct 9, 2004

### omicron

How did u come out with (11-b)(11+b)?

4. Oct 9, 2004

### Tide

I subtracted $b^2$ from both sides and factored the right side and the left side is a perfect square as you indicated.

5. Oct 9, 2004

### omicron

So I guess there is only one?

6. Oct 9, 2004

### Tide

What would that be?

7. Oct 9, 2004

### omicron

erm... 11?

Last edited: Oct 9, 2004
8. Oct 9, 2004

### Tide

If one is 11 then the other has to be 0 but you specified positive integers.

9. Oct 9, 2004

### omicron

Ok so how do i get it?
Btw i know that 0 isn't a positive integer, so what is it called then?

10. Oct 9, 2004

### TenaliRaman

if u cannot find b such that (11-b)(11+b) is a perfect square, what do u call such a situation?

0 is simply called zero.
However some literature do put zero in both positive and negative and indicate them separately as +0 and -0.+0 is positive zero and -0 in negative zero.

-- AI

11. Oct 9, 2004

### omicron

I don't know what u call such a situation.
Anyway back to the question. Do u know how many pairs of positive integer a, b are such that $$a^2+b^2=121$$

12. Oct 10, 2004

### TenaliRaman

a = (11-b)(11+b)
i cannot find positive integer b such that a is a perfect square
therefore there is no solution for a and b.

-- AI

13. Oct 10, 2004

Ok. Thanks!