Find-four-values-of-n-satisfying-1-n-2000-and-2-n-n-2-mod-1024-

Question Number 111503 by Aina Samuel Temidayo last updated on 04/Sep/20

Find four values of n satisfying

1 ⩽ n ⩽ 2000 and 2^{n} = n^{2} (\mod 1024)

Answered by 1549442205PVT last updated on 04/Sep/20

2^{n} = n^{2} (\mod 1024) \Leftrightarrow 2^{n} - n^{2} = k . 2^{10}

\Leftrightarrow n^{2} = 2^{n} - k . 2^{10} = 2^{10} (2^{n - 10} - k) (k \in N^{*})

\Leftrightarrow {(\frac{n}{32})}^{2} = 2^{n - 10} - k . \Rightarrow n = 32 m (m \in N^{*})

1 < n ⩽ 2000 \Rightarrow 32 m < 2000 \Rightarrow 1 ⩽ m ⩽ 62

\Rightarrow m^{2} = 2^{32 m - 10} - k (1)

m = 1 \Rightarrow n = 32, k = 2^{22} - 1

m = 2 \Rightarrow n = 64, k = 2^{54} - 4

m = 3 \Rightarrow n = 96, k = 2^{86} - 9

m = 4 \Rightarrow n = 128, k = 2^{118} - 16

Thus, we found four values of n

satisfying 2^{n} = n^{2} (\mod 1024) are

n \in {32, 64, 96, 128}

Commented by Aina Samuel Temidayo last updated on 04/Sep/20

Thanks .