Quandaries and Queries
 

 

Who is asking: Student
Level: Secondary

Question:
Hi

I have this problem which I think is a bit more tricky than usual.
x and y are positive integers.

1/x + 1/y = 2/31

It is possible to see that if x=y then 31 is a solution. But is it true that there are no other solutions?

31 is prime number and (x+y)/(xy)=2/31. So xy cannot equal 31 but in some way has to equal a number 31*n which means that x+y has to equal n*(x+y), correct?

Can I argue for no other solutions? If more solutions can be found, how can I proceed?

 

 

Matt,

For (x+y)/xy to simplify to 2/31, one of the two variables has to be a multiple of 31, so say x = 31 z. But then, (31z + y)/31zy simplifies to 2/31, so y is also a multiple of 31, say y = 31w. Then the equation becomes (31z + 31w)/31z31w = (1/31)*((z+w)/zw) = 2/31, that is, (z + w)/zw = (1/z + 1/w) = 2. What are the solutions of this in integers?

Claude

After reading this solution, Bruno sent us the solution 1/496 + 1/16 = 2/31. This means that there is something wrong with my solution. Can you find out what it is?

Claude

 
 

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