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Is it possible to find all positive integer triplets $(x,y,z)$ satisfying the parametric equation :

$$x^2 + 2ax + y^2 + 2by = z^2 + 2cz$$

Here $a, b, c$ are fixed positive integers.

Community
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kiva2000
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  • For such equations the solutions are always there. The formula there. http://math.stackexchange.com/questions/794510/curves-triangular-numbers – individ Apr 26 '15 at 17:34

1 Answers1

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One way to simplify a quadratic Diophantine is to get rid of cross-terms by doing a linear substitution. In this example, it is easy. Given,

$$x^2+2ax+y^2+2by = z^2+2cz\tag1$$

Let,

$$x,\,y,\,z = u-a,\,v-b,\,w-c$$

and it transforms to the simpler,

$$a^2+b^2-c^2-u^2-v^2+w^2=0$$

Since $a,b,c$ are constants, we can set $a^2+b^2-c^2 = d$, hence,

$$d-u^2-v^2+w^2=0$$

Do the further substitution $v,\,w =p+q,\,p-q$ to get,

$$d-4pq-u^2 = 0\tag2$$

I'm sure you can take it from here. P.S. All integer solutions of $(1)$ can be expressed in the form of $(2)$.

Tito Piezas III
  • 60,745
  • $$u^2+4pq = d\tag2$$

    Could this be a part of the solution (from Zagier) ?

    $$(u+2q,~q,~p-u-q), if u < p-q \tag2$$ $$(2p-u,~p,~u-p+q), if p-q < u < 2p\tag2$$ $$(u-2p,~u-p+q,~p), if u > 2p\tag2$$

     – kiva2000 Jul 30 '16 at 16:54
  • w2 - v^2 = 4pk It's true if only w and v have same parity, othercase w2 - v^2 = 4pk + 1. We can also write u^2-a^2 = 4ef and c^2-b^2=4gh if u and a have same parity and c and b also. In this cas we have 4pk+4ef+4gh = 0 => pk + ef + gh = 0 – kiva2000 Jan 16 '20 at 03:24