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This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ with } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ is also a prime number.

This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ where } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ is also a prime number.

This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ with } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ it also is a prime number.

This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ with } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ is also a prime number.

This conjecture is included in the larger following one

$A = \sigma_2(n) \ \ \text{ with } \ \ 1 <= n < \infty$

with the same other conditions  :

If $ \ A \ $ last digit is $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer it also is a prime.

This conjecture is part of the larger conjecture:

$A = \sigma_2(n) \ \ \text{ with } \ \ 1 <= n < \infty$

with the same other conditions:

If $ \ A \ $ has a last digit of $ \ 2 \ $, and if $\sqrt{A - 1}$ is an integer, then $\sqrt{A - 1} $ it also is a prime number.