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I wanna know, for the equation below, how to:

  • Prove if there is always a natural root $x$ that makes $y$ natural
  • Get the lowest natural $x$ that makes $y$ natural

$$ x^2+8x-y^2=4n-16\quad\forall n\in\mathbb N $$

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Note that the equation can be rewritten as $$(x+4)^2 - y^2 = 4n \implies (x+y+4)(x-y+4) = 4n =2n \times 2$$ Hence, one solution is \begin{align} x+y+4 & = 2n\\ x-y+4 & = 2 \end{align} Hence, $$2x+8 = 2n+2 \implies x = n-3$$ $$2y = 2n-2 \implies y = n-1$$ Hence, $(n-3,n-1)$ is always a solution for a given $n$.

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