Find all pairs

tony123 · 2008-07-23 05:14:48

Find all pairs of positive integers

such that

is a prime

and

is a perfect square

mathsyperson · 2008-07-23 06:19:39

All the pairs are generated as follows:

Choose an odd prime, p. Then set a = [(p+1)/2]² and b = [(p-1)/2]².

p is odd, which means p+1 and p-1 are both even. Then dividing them by two leaves integers, and so a and b are both perfect squares. Hence, ab is also a perfect square.

Now, a-b = [(p+1)/2]² - [(p-1)/2]² = [(p² + 2p + 1) - (p² - 2p + 1)]/4 = 4p/4 = p, which was chosen to be prime.

Hence, these values of a and b fit the conditions.

Furthermore, no other pairs exist. For two positive integers to multiply to give a square, they must be of the form a = xy², b = xz². Then you get ab = x²y²z² = (xyz)².

Then a-b = x(y² - z²) = x(y+z)(y-z).
For a-b to be prime, at least two of x, y+z and y-z must equal 1 and the other must be prime.
y+z has a minimum value of 2, which restricts the others to equal one.

Therefore, a and b must themselves be perfect squares and have consecutive square roots. With these restrictions, pairs can be made only using the method above.

ZHero · 2008-07-23 22:58:51

Impressive work.. mathsyperson!
I'd please like to know "what logic did you apply to start with"??

mathsyperson · 2008-07-24 01:49:43

Thanks!

I kind of worked it out backwards to the way I wrote it there. I started with a = xy² and b = xz² and found what restrictions needed to be placed on x, y, z, then used that to come up with the first part.

Math Is Fun Forum

#1 2008-07-23 05:14:48

Find all pairs

#2 2008-07-23 06:19:39

Re: Find all pairs

#3 2008-07-23 22:58:51

Re: Find all pairs

#4 2008-07-24 01:49:43

Re: Find all pairs

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