Name: Prime Numbers From 1 to 100
Uploaded: 2019-07-19
Duration: 7 min 49 s
Description: The number 2 is the only even prime number. But, how do you find the prime numbers between 1 and 100? This video explains the easiest way to find prime numbers.

Question

Find all primes

p

for which the quotient

\frac{(2^{p - 1} - 1)}{p}

is a square.

Accepted Answer

For $p = 2$ the given quotient is not a square and so we can assume $p$ is an odd prime. Then by Fermat's Little theorem $2^{p - 1} - 1$ is divisible by $p$ . Suppose that for some integer $a$ ,

$2^{p - 1} - 1 = p a^{2}$ .

Since, $p$ is odd we have that

$(2^{\frac{(p - 1)}{2}} - 1) (2^{\frac{(p - 1)}{2}} + 1) = p a^{2}$

Both the factors on the $LHS$ are odd and hence they are relatively prime to each other which implies that $p$ divides exactly one of the two factors. Therefore either

$(2^{\frac{(p - 1)}{2}} - 1) = p x^{2}, (2^{\frac{(p - 1)}{2}} + 1) = y^{2}$

or

$(2^{\frac{(p - 1)}{2}} - 1) = x^{2}, (2^{\frac{(p - 1)}{2}} + 1) = p y^{2}$ .

Case $1$ : Suppose $(2^{\frac{(p - 1)}{2}} - 1) = p x^{2}, (2^{\frac{(p - 1)}{2}} + 1) = y^{2}$

But then $2^{\frac{(p - 1)}{2}} = (y - 1) (y + 1)$ for which the only solution is $y = 3$ i.e., $p = 7$ . When $p = 7$ observe that the given quotient is indeed a square.

Case $2 :$ : Suppose $(2^{\frac{(p - 1)}{2}} - 1) = x^{2}, (2^{\frac{(p - 1)}{2}} + 1) = p y^{2}$ .

Then if $p > 3$ we get that $x^{2}$ leaves remainder $3$ when divided by $4$ which is impossible. Thus $p$ can only be equal to $3$ ; when $p = 3$ also we observe that the given quotient is indeed a square.
Thus, the primes $p$ for which the quotient $(\frac{2^{p - 1} - 1}{p})$ is a square are $p = 3, 7$ .

Find all primes $p$ for which the quotient $\frac{(2^{p - 1} - 1)}{p}$ is a square.

Important Questions on Number Theory

Learn and Prepare for any exam you want !

Find all primes p for which the quotient 2p-1-1p is a square.

Important Questions on Number Theory

Learn and Prepare for any exam you want !

Find all primes $p$ for which the quotient $\frac{(2^{p - 1} - 1)}{p}$ is a square.