# Squared fractions.

1. Aug 14, 2009

### icystrike

1. The problem statement, all variables and given/known data

Show that $$1/72$$
cannot be written as the sum of the reciprocals
of the squares of two different positive integers.

2. Relevant equations

3. The attempt at a solution

Available solutions
1/8²-1/24²
1/9²+1/648
Therefore Proven.

2. Aug 14, 2009

### Дьявол

$$\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{72}$$

$$\frac{b^2+a^2}{a^2b^2}=\frac{1}{72}$$

$$72(b^2+a^2)=a^2b^2$$

Solve the equation and write here the solution.

3. Aug 14, 2009

### icystrike

How do i solve that equation?

4. Aug 14, 2009

### Дьявол

Move the terms from the left to the right side of the equation:

$$a^2b^2-72b^2-72a^2=0$$

Now factor out b2 or a2 and tell me what you got.

5. Aug 14, 2009

### icystrike

$$a^2=(72b^2)/(b^2-72)$$

6. Aug 14, 2009

### Дьявол

Ok. Now what "a" is equal to? What can you conclude from the final solution?

7. Aug 14, 2009

### icystrike

Thanks for your help. But the actual problem is to derive the available solution from the question given.

8. Aug 14, 2009

### HallsofIvy

Yes, that is what he is trying to show you how to do!

However, the problem, as you stated it was

"Show that 1/72 cannot be written as the sum of the reciprocals
of the squares of two different positive integers." (my emphasis)

You can't do that because, as you showed, it is not true.

9. Aug 14, 2009

### icystrike

I got it. Im sorry.

Now for part 2,
How can I write $$1/72$$ with reciprocals of the squares of
three different positve integers.

Last edited: Aug 14, 2009
10. Aug 14, 2009

### Staff: Mentor

Start by writing an equation that expresses this relationship.

11. Aug 14, 2009

### icystrike

okay. $$1/a^2+1/b^2+c^3=1/72$$
By studying the relationship of their factor,
The equation can be translated into :
$$1/x^2+1/(b^2)(x^2)+1/(c^2)(x^2)=1/72$$
Moreover,

$$b^2+c^2+1=x$$