My first post and hope it is not a stupid one and related to this thread:
Can there be 4 whole numbers a, b, c, d all greater than zero such that:
a^2 + b^2 = c^2 and b^2 + c^2 = d^2?
If there are- can one give an example. If none can one prove it is not possible?
ty in advance!(Clapping)
It can be shown that an order triple, (a, b, c) is a "Pythagorean triple", a^2+ b^2= c^2, if an only if they can be written in the form a= n^2- m^2, b= 2nm, and c= n^2+ m^2 for some numbers m and n. And, of course, if we want a, b, c integers, m and n must be integers also.
Originally Posted by Kobby
In order that (a, b, c) be a Pythagorean triple, they must be of the form a= n^2- m^2, b= 2nm, and c= n^2+ m^2 for some integers m and n. In order that (b, c, d) also be a Pythagorean triple, they must be of the form b= p^2- q^2, c= 2pq, and d= p^2+ q^2 for some integers p and q. That means we must have b= 2nm= p^2- q^2 and c= n^2+ m^2= 2pq. Adding those equations, m^2+ 2mn+ n^2= (m+n)^2= p^2+ 2pq- q^2 so that p^2+2pq- q^2 is a perfect square. p^2+ 2pq- q^2= p^2+ 2pq+ q^2- 2q^2= (p+q)^2- 2q^2 so p and q must be such that (p+q)^2- 2q^2 is a perfect square.
So the whole question reduces to "can x^2- 2y^2 be a perfect square for any integers x and y?"
Still 995, 100, 1005 are not pytagorial numbers- so still no solution.
The fact that there is a^2 + 2b^2 = d^2, doesnt give you a, b, c as pytagorial numbers.
What all you all written I figured already.
Are there any a, b, c, d out there?
This is an interesting problem. I believe that these equations have no solutions. Suppose that there is a solution. Then after dividing through by any common factor we may assume that (a,b,c) and (b,c,d) are both primitive pythagorean triples. Thus c and d are odd, so b is even and a is odd. Then there are co-prime integers m (even) and n (odd) such that and . Therefore .
Originally Posted by Kobby
I believe (I haven't checked it carefully) that the diophantine equation has no solutions in positive integers. The proof would be by Fermat's method of infinite descent (the same approach that Fermat himself used to show that has no solutions).
Write the equation as , and factorise the left side: . If a prime factor of the odd number n divides both of those factors then it also divides their sum and hence divides m. But m and n are co-prime, so that cannot happen. Therefore the factors and must be and (either in that order or in the opposite order), where s,t are co-prime and .
If and then . But , and so . But those two factors are again co-prime, and therefore they must both be squares, say and , where . Thus and . So we have a solution to the same pair of equations as for a,b,c and d, but with smaller numbers.
The other possibility is that and , but this leads to a similar conclusion.
This process can now be repeated, to find a sequence of smaller and smaller solutions. But you cannot have an infinite, strictly decreasing sequence of positive integers. So we have a contradiction, and the conclusion has to be that the equations have no solution.
Thanks really. In the wikipedia value you had mentioned they also refered to the subject: "There are no Pythagorean triplets in which the hypotenuse and one leg are the legs of another Pythagorean triple."
So a puzzle is solved.
Glad I came across this site.(Happy)