# pythagoras question

• Oct 23rd 2009, 11:56 PM
Kobby
pythagoras question
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?
• Oct 24th 2009, 06:02 AM
HallsofIvy
Quote:

Originally Posted by Kobby
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?

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.

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?"
• Oct 24th 2009, 06:50 AM
aidan
Quote:

Originally Posted by Kobby
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?

\$\displaystyle
a^2 + b^2 = c^2\$
\$\displaystyle
b^2 + c^2 = d^2
\$

replacing \$\displaystyle c^2\$ in second equation
\$\displaystyle
b^2 + a^2 + b^2 = d^2
\$

rearrange/simplify
\$\displaystyle
a^2 + 2b^2 = d^2
\$

\$\displaystyle
995^2 + 2(100^2) = 1005^2
\$
(There are others.)
• Oct 24th 2009, 08:47 AM
Kobby
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?
• Oct 24th 2009, 11:46 AM
Opalg
Quote:

Originally Posted by Kobby
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?

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 \$\displaystyle b=2mn\$ and \$\displaystyle c=m^2+n^2\$. Therefore \$\displaystyle d^2 = b^2+c^2 = 4m^2n^2 + (m^2+n^2)^2 = m^4+6m^2n^2+n^2\$.

I believe (I haven't checked it carefully) that the diophantine equation \$\displaystyle m^4+6m^2n^2+n^2=d^2\$ 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 \$\displaystyle x^4+y^4=z^2\$ has no solutions).

Write the equation as \$\displaystyle (m^2+3n^2)^2 - d^2 = 8n^4\$, and factorise the left side: \$\displaystyle (m^2+3n^2+d)(m^2+3n^2-d) = 8n^4\$. 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 \$\displaystyle m^2+3n^2+d\$ and \$\displaystyle m^2+3n^2-d\$ must be \$\displaystyle 2s^4\$ and \$\displaystyle 4t^4\$ (either in that order or in the opposite order), where s,t are co-prime and \$\displaystyle st=n\$.

If \$\displaystyle m^2+3n^2+d = 2s^4\$ and \$\displaystyle m^2+3n^2-d = 4t^4\$ then \$\displaystyle m^2+3n^2 = s^4+2t^4\$. But \$\displaystyle n=st\$, and so \$\displaystyle m^2 = s^4 - 3s^2t^2 + 2t^4 = (s^2-t^2)(s^2-2t^2)\$. But those two factors are again co-prime, and therefore they must both be squares, say \$\displaystyle s^2-t^2 = p^2\$ and \$\displaystyle s^2-2t^2 = q^2\$, where \$\displaystyle m=pq\$. Thus \$\displaystyle q^2+t^2=p^2\$ and \$\displaystyle t^2+p^2=s^2\$. 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 \$\displaystyle m^2+3n^2+d = 4s^4\$ and \$\displaystyle m^2+3n^2-d = 2t^4\$, 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.
• Oct 24th 2009, 01:45 PM
Kobby
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)