A number theory problem by Sudhanshu Bharadwaj

Number Theory Level 3

For n > 2 n>2 , let A = 2 n + 1 A=2^n+1 and B = 1 + 2 + 2 2 + 2 3 + . . . + 2 n − 1 B=1+2+2^2+2^3+...+2^{n-1} .

Can both A A and B B be primes?

No Yes Depends on the value of n

Sudhanshu Bharadwaj by Sudhanshu Bharadwaj , 19, India

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1 solution

B is the sum of GP and on simplifying,

B=(2^n)-1

Notice that, B, 2^n ,A are 3 consecutive integers and 2^n is not divisible by 3

So either A or B must be divisible by 3 (a#b#3 as n>2) .So at least one of them must be a composite number

0 Helpful 0 Interesting 0 Brilliant 0 Confused

B=1+2+2^2+2^3+...+2^(n-1),not 1+2^2+2^3+...+2^(n-1)

X X - 3 years ago

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