Question

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FIND n,
if
$2 {}^{200} - 2 {}^{192}.31 + 2 {}^{n}$
IS A PERFECT SQUARE.

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Answer 1

hey friend your answer ↓

2^200-2^198.31+2^n

by taking 192 common

=2^192[2^8-31+2^(n-192)]

=2^192(256-31+2^m) ---{n-192=m}

=2^198(225+2m^2)

2^198 is a perfect square

substitute m such that 225+2m^2 also be a perfect square.

m=6 can satisfy our need.

put m=6

m= n-192

6=n-192

n= 192+6

n= 198

hope it helps you!

Answer 2

=2^192[2^8-31+2^(n-192)]

=2^192(256-31+2^m)-{n-192=m}

=2^198(225+2m^2)

2^198 is a perfect sq

Subtitude the value in given eq

m=225

put m=6

m=n-192

6=n-192

n=192+6

n=198

Hope you like my answer