If $a\ge 1$ and $k = (2a-1)q+2a-2 $ then $k\ge q$

Question

As is an offshoot of this earlier MSE question I seek to prove that if $a\ge 1$ and $k = (2a-1)q+2a-2 $ then $k\ge q$

MY ATTEMPT

Suppose to the contrary that both $$k=(2a-1)q+2a-2, a \geq 1$$ and $$k=1$$ hold.

Then we obtain $$1=(2a-1)q+2a-2$$ $$3-2a=(2a-1)q$$ $$q = \frac{3-2a}{2a-1}.$$

Let $$f(a) = \frac{3-2a}{2a-1}.$$ Differentiating $f$ with respect to $a$, we obtain $$f'(a) = -\frac{4}{(2a-1)^2} < 0$$ which means that $f$ is a decreasing function of $a$. Since $a \geq 1$, then we obtain $$f(a) \leq f(1) = \frac{3-2(1)}{2(1)-1} = 1.$$

This implies that $$q = \frac{3-2a}{2a-1} \leq 1.$$ But since $q$ is the special prime, satisfying $q \equiv 1 \pmod 4$, then we have $q \geq 5$. This is a contradiction.

This proves that the condition $k=1$ is incompatible with the assumption "$k=(2a-1)q+2a-2$ with $a \geq 1$".

Here is my:

FINAL QUESTION: As I find my proof in the MY ATTEMPT section unsatisfying, can you think of an alternative proof for the same fact?

Answer 1

Since $2a−1\geqslant 1$ and $2a−2\geqslant 0$, we have $$k=(2a−1)q+2a−2\geqslant 1\times q+0=q$$

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Another way :

Since $k$ is a function of $a$, writing $k$ as $$k=(2q+2)a-q-2$$ we get $$k=(\underbrace{2q+2}_{\text{positive}})a-q-2\geqslant (2q+2)\times 1-q-2=q$$

Answer 2

Here is an alternative (and direct) proof.

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Suppose that $q^k n^2$ is an odd perfect number with special prime $q$ satisfying $k=(2a-1)q+2a-2$ with $a \geq 1$.

We want to show that $k \neq 1$. (In other words, we wish to show that $k > 1$, and therefore that $k \geq 5$, since $k \equiv 1 \pmod 4$ ought to hold for positive $k$.)

Since $k=(2a-1)q+2a-2$ with $a \geq 1$ by assumption, we derive $$k+1=(2a-1)q+(2a-1)=(2a-1)(q+1).$$

Since $a \geq 1$ is an integer, then $2a-1 \geq 1$ holds, so that $$(q+1) \mid (k+1).$$

This last divisibility constraint implies that $$q + 1 \leq k + 1$$ from which we infer that $$q \leq k.$$

However, since $q$ is the special prime, satisfying $q \equiv 1 \pmod 4$, then we know that $5 \leq q$ holds.

We therefore conclude that $$5 \leq q \leq k,$$ or, by transitivity, $$5 \leq k.$$

QED