# nt.quantity concept – On queer consummate numbers \$p^ok m^2\$ with particular prime \$p\$ satisfying \$m^2 – p^ok = 2^r t\$ – Part II Answer

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## nt.quantity concept – On queer consummate numbers \$p^ok m^2\$ with particular prime \$p\$ satisfying \$m^2 – p^ok = 2^r t\$ – Part II

(Preamble: We have the identical query in. posed MSE two weeks in the past with out getting an respond. We have subsequently posted it to MO within the await that it’ll breathe answered right here.)

The theme queer consummate numbers in all probability would not necessity an introduction.

Designate the traditional sum of the divisors the optimistic integer $$x$$ from $$sigma (x) = sigma_1 (x)$$.

if $$n$$ is peculiar and $$sigma (n) = 2n$$, then we convene $$n$$ a queer consummate quantity. Euler proved {that a} putative queer consummate quantity essentially takes the figure. will need to have $$n = p ^ km ^ 2$$ Where $$p$$ is the particular prime quantity well behaved $$p equiv ok equiv 1 pmod 4$$ and $$gcd (p, m) = 1$$.

Descartes, after which Frenicle Sorli guessed that $$ok = 1$$ all the time lasts. Dris guessed that the inequality $$p ^ ok is undoubted in his M.Sc. thesis, and Brown (2016) lastly supplied proof of the weaker inequality p .$$

Recent proof means that $$p ^ ok can truly breathe grievance.$$

THE ARGUMENT

Permit $$n = p ^ km ^ 2$$ breathe an queer consummate quantity with a particular prime quantity $$p$$.

Since $$p equiv ok equiv 1 pmod 4$$ and $$m$$ is peculiar then $$m ^ 2 – p ^ ok equiv 0 pmod 4$$. Aside from that, $$m ^ 2 – p ^ ok$$ is not a sq. (Dris and San Diego (2020)).

This implies that we’re allowed to jot down
$$m ^ 2 – p ^ ok = 2 ^ rt$$
Where $$2 ^ r neq t$$, $$r geq 2$$, and $$gcd (2, t) = 1$$.

To show that’s trifling $$m neq 2 ^ r$$ and $$m neq t$$in order that we deem the next instances:

$$textual content {Case (1):} m> t> 2 ^ r$$
$$textual content {Case (2):} m> 2 ^ r> t$$
$$textual content {Case (3):} t> m> 2 ^ r$$
$$textual content {Case (4):} 2 ^ r> m> t$$
$$textual content {Case (5):} t> 2 ^ r> m$$
$$textual content {Case (6):} 2 ^ r> t> m$$

We can simply rule the illustration out (5) and illustration (6), as follows:

Under housing (5), we’ve got $$m and m <2 ^ r what implies that m ^ 2 <2 ^ rt . Which offers 5 leq p ^ ok = m ^ 2 – 2 ^ rt <0, which is a contradiction.$$

Under housing (6), we’ve got $$m <2 ^ r$$ and $$m what implies that m ^ 2 <2 ^ rt . Which offers 5 leq p ^ ok = m ^ 2 – 2 ^ rt <0, which is a contradiction.$$

Under housing (1) and illustration (2), we will show the inequality $$m$$

applies as follows:

Under housing (1), we’ve got:
$$(m – t) (m + 2 ^ r)> 0$$
$$p ^ ok = m ^ 2 – 2 ^ rt> m (t – 2 ^ r) = m left | 2 ^ r – t privilege |.$$

Under housing (2), we’ve got:
$$(m – 2 ^ r) (m + t)> 0$$
$$p ^ ok = m ^ 2 – 2 ^ rt> m (2 ^ r – t) = m left | 2 ^ r – t privilege |.$$

So remain with Case now (3) and illustration (4).

Under housing (3), we’ve got:
$$(m + 2 ^ r) (m – t) <0$$
$$p ^ ok = m ^ 2 – 2 ^ rt$$

Under housing (4), we’ve got:
$$(m – 2 ^ r) (m + t) <0$$
$$p ^ ok = m ^ 2 – 2 ^ rt$$

Notice that beneath illustration (3) and illustration (4), we even have
$$min (2 ^ r, t)$$

But the situation $$left | 2 ^ r – t privilege | = 1$$ sufficient for $$p ^ ok retain.$$

Our request is:

QUESTION: Is the situation $$left | 2 ^ r – t privilege | = 1$$ too needful for $$p ^ ok maintain, beneath illustration (3) and illustration (4)?$$

Notice the situation $$left | 2 ^ r – t privilege | = 1$$ contradicts the inequality
$$min (2 ^ r, t) beneath the remaining illustration (3) and illustration (4), and the truth that m is an integer.$$

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