Integer is easy

Level 1

Find the sum of all integer solutions of

$2^n + 1 = n^2$

for $n > 1$

none of these 99 2018 138 3

1 solution

Atishay Jain
Dec 3, 2017

Given, $2^n = (n+1)(n-1)$

Now, simply n has to be odd.

Let say if n+1 is multiple of any number other than a number like, $2^1, 2^2, 2^3,....$ , for eg. 6, then in order to cancel the other factor i.e 3 in this case n-1 has to be a rational with non-one denominator in this case is 3.

Therefore, n + 1 as well as n - 1 has to be a number of type $2^1, 2^2, 2^3....$ .

There, exists only one such pair with difference 2 i.e. 2 & 4. This implies that n - 1 = 2 & n + 1 = 4.

Therefore, n = 3 only one integer solution

Integer is easy

1 solution

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