Inverse Pythagorean Theorem
Find the number of integer solutions for p , q , and r that satisfy the equation:
p 2 1 + q 2 1 = r 2 1
where there is no common integer factor greater than 1 for all three variables p , q , and r .
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2 solutions
Oops, I didn't think about that. I'm going to change the wording so that p, q, and r have no common factor.
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Getting ready to edit my solution...
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The second part is very clever and makes a good solution to my edited problem! I think there is a typo, though: r = ab, not r = bc.
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@David Vreken – Thanks for pointing out the typo! I've updated the solution now.
I am obliged to point out that any set of positive integers always have a common factor of 1.
Primitive inverse Pythagorean triplet Chris? :) I was about to post a solution and seen yours and controlled myself. :p
What if all the three did not contain any common factor. Is it even solvable? I mean, p = 2 , q = 3 , r = 5 something like that. All of them being co-prime to each other
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Oh, I meant to add this to my solution.
Imagine you have a triple ( p , q , r ) that satisfies the equation. If you multiply through by ( lcm ( p , q , r ) ) 2 , it becomes a primitive Pythagorean triple. So the argument is reversible, and p , q , r can't be pairwise coprime.
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Wow, I didn't know about that. Can you elaborate on why this must be true?
We could rewrite it to
p 2 r 2 + q 2 r 2 = p 2 q 2
Which is similar to a 2 + b 2 = c 2
We know infinitely many Pythagorean triplets exist
Therefore, there are infinite solutions for ( p , q , r )
You would also need to show that there are infinitely many Pythagorean triplets in the form of (pr, qr, pq).
I made the same equation, and clicked "None" :) But the question is good.
Yeah you need to prove there exist triplets whose sides are not prime and are of the form, pr, rq and pq.
But you need to show that (pr,qr,pq) can form a Pythagorean triplet and there are infinite of them.
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Yes, but (pq, qr, rp) are independent of each other. So, they are equivalent to (a, b, c), isn't it? :)
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If such p,q,r exist which satisfy the equation, (1/p^2)+(1/q^2)=(1/r^2) then such a Pythagorean triplet (pq, qr, rp) exist, but here you have to show that (1/p^2)+(1/q^2)=(1/r^2) has p,q and r as integer solutions with no common integer factor greater than 1, so you need to show that (pq, qr, rp) exist to show as we can reverse our process.
If p^2 + q^2 = (pq/r)^2 and p and q are integers, then pq/r is also an integer. So, pq must be divisible by r. But the question says, there is no common integer factor greater than 1 for all three variables p, q, and r (I assume this mean, p, q and r are pairwise co-primes). Doesn't this mean there is no such triplet (p, q, r)? Where did I go wrong?
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If p is divisible by r and q is not divisible by r, then pq/r is an integer and there is no common integer factor greater than 1 for all three variables p, q, and r.
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Oh! I thought gcd(p, q) = gcd(q, r) = gcd(r, p) = 1 instead of, gcd(p, q, r) = 1 Thanks for clafrifying it.
Take any primitive Pythagorean triple, say a 2 + b 2 = c 2 . Now divide through by ( a b c ) 2 to get ( b c ) 2 1 + ( c a ) 2 1 = ( a b ) 2 1
so that p = b c , q = c a and r = a b is a solution to the given equation. Since there are infinitely many primitive Pythagorean triples, this gives infinitely many solutions.
For example, with ( a , b , c ) = ( 3 , 4 , 5 ) , we get 1 5 2 1 + 2 0 2 1 = 1 2 2 1