Prove that $\sqrt{(p-1)^2 p^2 + 4} = (p - 1)p + \epsilon$ for $p > 1$ and $0 < \epsilon < 1$.

Question

Here $p$ is prime but that does not affect the calculation. Assume that $p > 1$ is any such integer. Numerically at $p = 2$ we have $\epsilon = 0.8...$ which is the maximum value of $\epsilon$.

Answer 1

You want to show that

$$\sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$

Because both sides are guaranteed to be positive, we can square both sides with no problem

$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$

Subtracting

$$4 < 2p(p-1)+1$$

Rearranging

$$0 < 2p^2-2p-3$$

The quadratic on the RHS has zeroes at $\approx -0.8$, and $\approx1.8$, and it opens up. Therefore for $p \ge 2$, the inequality holds. And this inequality is equivalent to your original one.

EDIT:

Oh and also you can show that $\epsilon$ has to be positive because $p(p-1) = \sqrt{p^2(p-1)^2} < \sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.

Answer 2

You have the inequality $$\sqrt{1+x}\le 1+\frac x2$$

This can be shown easily since $1+x\ge 0\implies 1+\frac x2>0$ so we can compare their squares and this give $1+x\le 1+x+\frac{x^2}4$

Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<\sqrt{y^2+a}\le y+\frac a{2y}$$

So the question is to find if $\varepsilon=\dfrac{a}{2y}<1\iff y>\dfrac a2$

For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$

and for $p=2$ just verify manually that $\sqrt{8}-2\approx 0.82<1$.