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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

Recent proof means that $ p ^ ok

**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

$$ 5 leq p ^ ok = m ^ 2 – 2 ^ rt <0, $$

which is a contradiction.

Under housing **(6)**, we’ve got $ m <2 ^ r $ and $ m

$$ 5 leq p ^ ok = m ^ 2 – 2 ^ rt <0, $$

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

Our request is:

QUESTION:Is the situation $ left | 2 ^ r – t privilege | = 1 $ too needful for $ p ^ okmaintain, 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.

we’ll proffer you the answer to 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 query by way of our community which brings all of the solutions from a number of dependable sources.