The solution of the Diophantine equation x^2 +y^2=z^2 (Pythagorean triples) are x=2mn , y=m^2 -n^2, z=m^2 + n^2.
We look for the integer solutions of the Diophantine equation (xy)^2+(yz)^2+(zx)^2=(pq)^2 in polynomial forms. x, y, z are expressed in polynomial forms with integer variables n,m, l in a way similar to Pythagorean triples.
x=A(n,m,l), y=B(n,m,l), z=C(n,m,l), p=D(n,m,l), q=E(n,m,l)
The degre of polynoms A, B, C,D,E is less than or equal to 4.
Should p and q be prime? If not, we can just use one variable r=pq instead of two.
Dear Marco
Thank you for your answer. Two cases:
1/ p and q are prime;
2/ p and q are not prime then we can use r=pq
Sincerely
What about the following solution:
x=m(m2-k2+l2), y=l(m2-k2+l2), z=2klm
giving
(xy)2+(yz)2+(xz)2=d2
with d=lm(k4-(l2+m2)2)
?
Probably all solutions have a multiple that is of this form up to sign and sequence.
Dear Dieter, your solution works well!
Do you have some arguments for your last statement, or is this an intuition?
Dear Dieter
Your solution is ok in the case of the right member is a perfect square but what about if the right member is pq?
The variables x; y and z are similar but this characteristic is not visible in your solution
Thanks a lot
I found my solution like this: it is easy to see that every rational solution x, y, z is of the form
y = b x, z = 2xab/(1-a2+b2) with some rational a and b, x any rational.
By setting b = m/k, a = l/k with GCD(k,i,m)=1 and clearing denominators I find a multiple of (x,y,z) of the form given in my previous answer and consisting of integers. So every rational solution has a multiple of my form and of integers, so this is true for integer solutions also. But not all integer solutions are of this form, e.g. (3,1,2) , but its multiple (18,6,12) is with (k,l,m)=(1,2,3).
My solution tends to have large divisors in common, but I had not enough patience and scrutinity to investigate common divisors.
Dear Mohamed, I don't understand your last question. If the rhs isn't a perfect square, the answer to the equation is trivial. Or are you looking for squared primes as rhs?
Yes, x,y,z are similar, and this is not visible, but that is quite common, even for pythagorean triples m2-n2, 2mn, m2+n2.
Dear Dieter
If we choose l=m=n=1 we get x=1, y=1, z=2 and d= -3
It seems that your solution works algebricaly but not arithmeticaly, I means in Diophantine equation we have to find positive integers.
My intuition we have to look for x, y and z as a polynom of degre 4 with l,m,n as variables.
x, y and z are intrerchangeable and related (xy , yz, zx) then the solution must be similar and interchangeable but d is different. This Diophantine equqtion is different from
x^2 +y^2=z^2
Yes there is a case where we are looking for squared primes as rhs.
Thank you
Dear Marco,
Thank you for your answer but it does not fulfil the requirments. We look for a polynomial solution with 3 independent variables l,m,n. The degre of polynom is less than or equal 4. In addition x, y, z are independent variables. Here z=x+y
The answer could be the solution for y^2 + z^2 + (yz)^2 = r^2
Regards
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