Help Needed: Solving Integer Equations with GCD = 1 - Angelo Spina

[angelo]

I am a visitor of this beautiful site, my name is Angelo Spina, I would like to resolve the three following problems, in fact after many attempts I have not succeeded in it, for this reason I kindly ask you to give me a help.


PROBLEM 1.

If the equation y² + a p² = 2 x² (where a is a positive integer, p is an odd prime number) admits a solution (y,x) of integers with gcd(y,x)=1, how can I prove that the equation y² + p² = 2 x² also admits a solution (y,x) of integers with gcd(y,x)=1 ?


PROBLEM 2.

If the two equations y² + p² = 2 x² , y² + q² = 2 x² (where p and q are odd prime numbers) respectively admit the integer solutions (y',x') and (y '', x '') with gcd(y',x')=1 and gcd(y'',x'')=1, how can I prove that the equation y² + (pq)² = 2 x² also admits at least an integer solution (y*,x *) with gcd(y*,x*)=1 ? Is it possible to find a formula that allows to obtain (y*,x *) from the knowledge of (y',x') and (y '', x '') ?


PROBLEM 3.

If the equation y² + n² = 2 x² (where n is a positive integer greater than 3) admits integer solutions con gcd = 1, how can I prove that the equations
y² + p² = 2 x², y² + q² = 2 x², y² + r² = 2 x², y² + s² = 2 x²,..., (where p, q, r, s,...are the prime factors of n) also admit integer solutions con gcd=1?


Certain of your courtesy, I thank you very much.

Angelo.

 

[Zurtex]

Hello and Welcome

Please can you tell us any work you have been able to do on this so far.

 

[matt grime]

May I ask where these question arise from? If it's a basic course in number theory and you've done certain topics presumably these topics are of use in solving the problem (is there a similar problem in the notes/book that is solved)? Roughly what knowledge does this require to solve. If this is a problem from a class field theory text nook then we almost certainly won't know the answer. If its taken from a textbook at a level where observations such as 2 is a quadratic residue mod p^2 and a is a residue mod x^2 and 2x^2 are what the other proofs in the book/notes near to this one are involved then we possibly can help.

 

[ramsey2879]

The solution starts with the simple identity that if Z is an odd square then (Z-1)/8 equals a triangular number, say A(A+1)/2. Set A(A+1)/2-A(A+1) = -A(A+1)/2 and make the substitution A = (X-1)/2. Multiply both sides of the resulting equation by 8 and substract 1 from each side gives Y^2-2X^2=-Z^2 or Y^2 + Z^2=2*X^2. This solves problems 2 and 3.

 

[ramsey2879]

I posted the message on 7/2 before leaving to visit relatives. I then realized that my answer gave gcd > 1. I didn't have access to a computer but tried to work the "problem" with using a simple calculator. Unfortunately I used the equation y^2 + 2*x^2 = P^2 instead of y^2 + P^2 = 2*x^2. I will save this work for a separate post. As to the current problem I think the relation 2((x-y)^2+x^2)=(2x-y)^2 + y^2 will help. Simply use the well known relation for Pythagorean triples to get (x-y)^2 + x^2 = X^2 and set P= 2x-y this will give a general equation for solutions to Y^2 + P^2 = 2X^2 but I don't know if it is inclusive of all solutions.