## Prime number 3

$3^4 = 81$   has 5 divisors:     1,   3,   9,   27,   81

Sum of divisors:     $1 + 3 + 9 + 27 + 81 = 121 = 11^2$

Are there any other primes   $p$   such that the sum of all positive integer divisors of   $p^4$   is equal to a square of an integer?

Solution:

If   $p$   is a prime, then the sum of all positive integer divisors of   $p^4$   equals

$1 \; + \; p \; + \; p^2 \; + \; p^3 \; + \; p^4$

If $1 \; + \; p \; + \; p^2 \; + \; p^3 \; + \; p^4 \; = \; n^2$

where   $n$   is a positive integer.

then we have obviously

$(2 \,p^2 + p)^2 \; < \; (2 \,n)^2 \; < \; (2 \,p^2 + p + 2)^2$,

and it follows that we must have

$(2 \,n)^2 \; = \; (2 \,p^2 + p + 1)^2$

Thus,

$4 \, n^2 \; = \; 4 \, p^4 \; + \; 4 \, p^3 \; + \; 5 \, p^2 \; + \; 2 \, p \; + \; 1$

and since

$4 \, n^2 \; = \; 4 \, (p^4 + p^3 + p^2 + 1)$

we have

$p^2 \; - \; 2 \, p \; - \; 3 \; = \; 0$

which implies     $p \; | \; 3$

hence     $p \; = \; 3$

If fact, for   $p \; = \; 3$   we obtain   $1 + 3 + 3^2 + 3^3 + 3^4 = 11^2$

Thus, there exists only one prime   $p$,   namely   $p \; = \; 3$.