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July 26

Diophantine Equation

Let a be some natural number. What are the solutions in integers of the equation $x^2+4a=y^2$? עברית (talk) 16:32, 26 July 2017 (UTC)

Any even difference of 2 squares is divisible by 4, so there are many solutions to that equation! Georgia guy (talk) 16:38, 26 July 2017 (UTC)

Template:Em: Try rewriting as $y^2-x^2=4a$, and then factor and solve in terms of the factors of 4a. Considering whether the factors are even or odd should let you simplify further. --Deacon Vorbis (talk) 16:47, 26 July 2017 (UTC)

• Template:Ec Some random thoughts:

Consider a prime number p≠2 that divides both x and y, then p² must divide a since it divides 4a and not 4, and then (x/p, y/p) is solution for a/p². Hence, the general solution should be easily generated from the solutions where $GCD(x,y)=2^k$ by stripping a from its square divisors.

For the divisibility by 2: Obviously x and y are both odd or both even. If they are even, the equation reduces to $a=(y^{\prime }-x^{\prime })(y^{\prime }+x^{\prime })$ where $2x^{\prime }=x,2y^{\prime }=y$. If they are odd, $4a=y^2-x^2=(2y^{\prime }+1{)}^2-(2x^{\prime }+1{)}^2=4y{\text{'}}^2-4x{\text{'}}^2+4y^{\prime }-4x^{\prime }$ which reduces to $a=(y^{\prime }-x^{\prime })(y^{\prime }+x^{\prime }+1)$. Both equations look quite attackable by looking at the pairs m,n such that a=m*n, and retrieving x,y as a function of m,n. Tigraan^{Click here to contact me} 17:24, 26 July 2017 (UTC)

A curious fact is that the equation always has at least two solutions $x=a-1$, $y=a+1$ and $x=-a+1$, $y=-a-1$. These are the only solutions if $a$ is simple. Ruslik_Zero 17:32, 26 July 2017 (UTC)

Thank you! עברית (talk) 19:40, 26 July 2017 (UTC)