Expert: David Hemmer - 1/2/2011

Question
Hello Mr. Hemmer,

I am a 14 year old in high school and I was practicing for the David Essner math finals by trying previous test questions. The solutions aren't given online, so I was wondering if I did the following question correctly: (I only answered parts A and C because parts B and D were similar except they used composite numbers)

The Reciprocals Triple Problem:
In this problem x,y,z will denote positive integers where x>1, y>1, z >= 1 and 1/x + 1/y = 1/z. An example of an (x,y,z) triple for this equation is (2,2,1) since 1/2 + 1/2 = 1/1.

a) Part 1: For x=3, find a triple (x,y,z) such that 1/x + 1/y = 1/z.
Part 2: For x=7, find a triple (x,y,z) such that 1/x + 1/y = 1/z.

c) Prove that if x is a prime number then there is a unique [only one] triple (x,y,z) such that 1/x + 1/y = 1/z.

My answer:
Look at part c first.

c) Because 1/x + 1/y = 1/z, 1/y = 1/z - 1/x = (x-z)/(xz). Let n denote the difference (x-z). Then 1/y = n/[x(x-n)]. For y to be an integer, the fraction n/[x(x-n)] would have to simplify to make n=1. One triple, then would occur when n already equals 1 (i.e. x is exactly 1 greater than z). When x is prime, there are no other solutions. For there to be other solutions where x is prime, the fraction n/[x(x-n)] would not simplify to give n=1 since neither n/x nor n/(x-n) will simplify because: x and n will be relatively prime since x is prime, and n and (x-n) can be shown to be relatively prime in the following way: (1/x)*[n/(x-n)] = 1/y. We "simplify" to make the n/(x-n) numerator 1 and get (1/x)*[1/([x/n]-1)] = 1/y. Then y = [(x^2)/n]-1. But y will then not be an integer because (x^2)/n will not simplify to an integer (because x is prime). Therefore, given a prime x, there is only one solution triple, with z = x - 1 and y = x(x-1).

a) Part 1: Using z = x - 1 and y = x(x-1), we get z=2 and y=6 to make the triple (3,6,2)
Part 2: Using z = x - 1 and y = x(x-1), we get z=6 and y=42 to make the triple (7,42,6).

Here are the questions in parts (b) and (d) if you'd like to see them:

b) Determine for each of (Part 1) x=6 and (Part 2) x=9 two triples (x,y,z) such that 1/x + 1/y = 1/z.

d) Prove that if x is not a prime number then there are at least two triples (x,y,z) such that 1/x + 1/y = 1/z.

Thank you, any feedback will be appreciated.

Answer
Your proof has a little confusion about relatively prime. Relatively prime means the GCD is 1, so you don't automatically know x and n are relatively prime, it could be n is a multiple of x. Instead I might take your equation:

ny=x(x-n) and put the y and z back in (i.e. substitute n=x-z). You get:

(x-z)y=-xz

which gives:

x(z+y)=zy

Now since x is prime and it goes into the left it also divides zy, so zy is a multiple of x. Now this is the key property of prime numbers (indeed this is how primes are defined in more abstract settings.) Namely, if zy is a multiple of x, then either z or y is.

Well x is larger than z so we can assume y is a multiple of x. Try to show why this can't happen.