615 + x^2 = 2^y ; x;y∈N ⇒ x;y = ?


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Question Number 148947 by mathdanisur last updated on 01/Aug/21

$615 + x^{2} = 2^{y}; x; y \in N$ $\Rightarrow x; y = ?$
	Answered by dumitrel last updated on 02/Aug/21
	$if y=2k+1⇒u(2^y )∈{2,8}⇒ u(x^2 )∈{3,7}⇒imposibil ⇒y=2k⇒615=(2^k −x)(2^k +x)⇒imposibil 2^k −x=5 2^k +x=123⇒k=6,x=59 ⇒x=59,y=12$
	$i f y = 2 k + 1 \Rightarrow u (2^{y}) \in {2, 8} \Rightarrow$ $u (x^{2}) \in {3, 7} \Rightarrow i m p o s i b i l$ $\Rightarrow y = 2 k \Rightarrow 615 = (2^{k} - x) (2^{k} + x) \Rightarrow i m p o s i b i l$ $2^{k} - x = 5$ $2^{k} + x = 123 \Rightarrow k = 6, x = 59$ $\Rightarrow x = 59, y = 12$
	Commented by mathdanisur last updated on 01/Aug/21

	$T h a n k Y o u S e r, C o o l$
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