−p^2 +2027=−q^2 p+q=?


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Question Number 90013 by ar247 last updated on 20/Apr/20

$- p^{2} + 2027 = - q^{2}$ $p + q = ?$
Commented by mr W last updated on 21/Apr/20

$p^{2} - q^{2} = 2027 \leftarrow p r i m e$ $(p - q) (p + q) = 1 \times 2027$ $f o r p, q \in N$ $p + q = 2027$ $p - q = 1$ $\Rightarrow p = 1014, q = 1013$
Commented by MJS last updated on 21/Apr/20

$for your solution, u = \frac{2027}{2}$
	Answered by MJS last updated on 21/Apr/20
	$p^2 −q^2 −2027=0 p=u−v∧q=u+v −4uv−2027=0 v=−((2027)/(4u)) p=u+((2027)/(4u))∧q=u−((2027)/(4u)) p+q=2u with u≠0 ⇒ p+q=z; z∈C\{0}$
	$p^{2} - q^{2} - 2027 = 0$ $p = u - v \land q = u + v$ $- 4 u v - 2027 = 0$ $v = - \frac{2027}{4 u}$ $p = u + \frac{2027}{4 u} \land q = u - \frac{2027}{4 u}$ $p + q = 2 u with u \neq 0$ $\Rightarrow p + q = z; z \in C ∖ {0}$
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