describe odd number calculation

Question

If a odd number 35

two number addition 26 + 9 = 35

here small number 9 * 3 = 27

27 - 26 = 26 - 27 = +1 or -1

here 35 is the one of the odd number , from the odd number 35 which two number addition to get number 35 , and the two number full fill the above calculation , numbers are 26 and 9 addition to get 35 , small number 9 mul by 3 get 27 , small number is 9 and other big number 26 , (big number minus small number mul by 3) or (small number mul by 3 minus big number) to get ans +1 or -1

another number 37

two number addition 28 + 9 = 35

here small number 9 * 3 = 27

28 - 27 = 27 - 28 = +1 or -1

another number 43

two number addition 32 + 11 = 43

here small number 11 * 3 = 33

33 - 32 = 32 - 33 = +1 or -1

and like above odd number is 55 how to get two number like above calculation any formula or method

two number addition ? + ? = 55

here small number ? * ? = ?

? - ? = +1 or -1

here how to get those two number , and those two number full fill the above calculation any formula or method need

Answer 1

It was very hard to figure out what, if anything you were asking.

I finally figured you meant

$a + b = 55$ where $b $ is odd.

$b*c = a-1$

So $a-b*c=1$

Okay, we just solve it.

$a=55-b$

$b*c=54-b $

$b*(c+1)=54$

So $b $ can be $1,3,9,27$.

$54 +1 =55;1*53=53;54-53=1$

$52+3=55;3*17=51;52-51=1$

$46+9=55;9*5=45;46-45=1$

$28+27=55;27×1=27;28-27=1$

But, geez, it was practically impossible to guess what the heck you were trying to ask. It seemed like you were completely making everything up.

I was going to answer just do $48+7=55; 3*6=18;17-16=1$. That made about as much sense as your question seemed to.

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More answers. Seeing your latest post it seems you want:

$a + b =55; a < b;a \le 22 $

$a*c = b \pm 1$

So $a*c =55 \pm 1 - a $

$a*(c+1)=54,56$

So $a= 1,2,3,4,6,7,8,9,14,18, $

$b=54,53,51,etc. $

And $c=53,55,26,27,etc... $

Answer 2

Your problem is equivalent to finding a solution $(x,y)$ with $x < y$ of the following system: $$\left\{ \begin{matrix} x&+&y& =& 55 \\ 3x && \pm 1 & =& y\end{matrix} \right.$$ if you sum these two equations side by side, you can get an equation just with $x$ $$\left\{ \begin{matrix} 4x&& \pm 1& =& 55 \\ 3x && \pm 1 & =& y\end{matrix} \right.$$ whose unique integer solution is $(x,y)=(14,41)$.