II. Conditions of resolvability of the equation
in integers.

We solve the equation
using a new method of the solution of the equations of Eremin (see M.A.Eremin “A new method of the solution of the equations” Arzamas 2000).

So
, where
,
- integers, 





Let's write down the solution of the equation
:


Let's put
, then:


If to take
,



Let's inspect the decision:


Thus
.

So, formulas
give the solution of the equation
.


In order that the equation
had the decision in integers, it is necessary that
,
were the whole positive numbers at
integers.




Let




From
we will receive:



From here we find
, from equality
we find
.



So:
, 
are criteria of resolvability of the equation
in integers.




*u-can have an infinite number of integer solutions.
If
numbers), the equation |
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Formulas
of the solution of the equation
can be written down in a look:



If the number
nonintegral, the solution of the equation
is ambiguous.


Consequence 1:
From
follows, if
, and
, the equation
has the following decision:





Consequence 2:
From
follows, if
, and 
:





Consequence 3:
From
follows, numbers
surely have the general multiplier.


The general multiplier can accept the following values:

Let's show it.
Numbers
for the equation
are equal:



а) Let
, then
. Let 
, then
.





So, if 
, the general multiplier of numbers
is equal




If 
, that 
.




So, if
, and
, numbers
have the general multiplier




б) Let
, then
. If
, that
.





Thus, if
, and
, the general multiplier of numbers
is equal




Let

, then, or
. In this case general multiplier of numbers




it is equal
.

в) Let




If
. So, if
, and
, the general multiplier of numbers
is equal:






If
, that
, in this case the general multiplier of numbers
is equal
.




Consequence 4:
From
follows, numbers
surely have the general divider:


а)
, where
such that
was an integer;



б)
;

в)
- if number
, where
- an integer;



г)
- if number
, where
- an integer.



The general divider of numbers
can be any natural number more unit.

The general divider of numbers
can be both simple, and compound number.

At
,
- simple number, minimum possible.


The equation
if
does not allow
to make in numbers reduction on their general divider. At reduction the equation essence is broken.



Let's show it:

Let
, where
- the whole positive numbers.











