Is p^n Deficient if p is a Prime?

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Homework Statement



Show that if p is a prime, then p^n is deficient.

Homework Equations





The Attempt at a Solution



I have no idea where to start.
 
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You could start with the definition of a deficient number.
If p is prime, it's only divisors are 1 and p and thus, the sum of it's divisors is p+1.
What are the divisors of pn?
 
VeeEight said:
You could start with the definition of a deficient number.
If p is prime, it's only divisors are 1 and p and thus, the sum of it's divisors is p+1.
What are the divisors of pn?

would be P^n +P^1+P^0 correct?
 
If p is prime, then it's only divisors are 1 and p. The divisors of pn are also 1 and p, which sum to p+1. Apply this to the criteria of a deficient number.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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