how many positive x≤10 000 integers are such that 2^x −x^2 is divisible by 7?


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Question Number 154672 by talminator2856791 last updated on 20/Sep/21

$how many positive x ⩽ 10 000 integers are$ $such that 2^{x} - x^{2} is divisible by 7 ?$
	Answered by MJS_new last updated on 20/Sep/21
	$2^(3n) =7k+1 2^(3n+1) =7k+2 2^(3n+2) =7k+4 (7n)^2 =7k+0 (7n+1)^2 =7k+1 (7n+2)^2 =7k+4 (7n+3)^2 =7k+2 (7n+4)^2 =7k+2 (7n+5)^2 =7k+4 (7n+6)^2 =7k+1 ⇒ n=0, 1, 2, 3,...: (2^n −n^2 )/7=7k+ 1 1 0 6 0 0 0 2 3 4 0 2 4 1 4 0 5 2 6 5 3 1 1 0 6... ⇒ remainders are 0 for n=21k+{2, 4, 5, 6, 10, 15} 21×476+5=10001 ⇒ ⇒ answer is 475×6+2=2852$
	$2^{3 n} = 7 k + 1$ $2^{3 n + 1} = 7 k + 2$ $2^{3 n + 2} = 7 k + 4$ ${(7 n)}^{2} = 7 k + 0$ ${(7 n + 1)}^{2} = 7 k + 1$ ${(7 n + 2)}^{2} = 7 k + 4$ ${(7 n + 3)}^{2} = 7 k + 2$ ${(7 n + 4)}^{2} = 7 k + 2$ ${(7 n + 5)}^{2} = 7 k + 4$ ${(7 n + 6)}^{2} = 7 k + 1$ $\Rightarrow$ $n = 0, 1, 2, 3, . . . : (2^{n} - n^{2}) / 7 = 7 k +$ $1 1 0 6 0 0 0 2 3 4 0 2 4 1 4 0 5 2 6 5 3 1 1 0 6 . . .$ $\Rightarrow$ $remainders are 0 for$ $n = 21 k + {2, 4, 5, 6, 10, 15}$ $21 \times 476 + 5 = 10001 \Rightarrow$ $\Rightarrow answer is 475 \times 6 + 2 = 2852$
	Commented by mathdanisur last updated on 20/Sep/21

	$very nice$
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