Question

$\bf \: find \: \alpha \: and \: \beta \: satisfying \: \mapsto3 \alpha + 2 \beta = \alpha \beta + 1.$
●need detailed solution!
​

Answer 1

[Note: Here $\alpha$ is denoted by $x$ and $\beta$ is denoted by $y.$ ]

Complete question:

$3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+.$

Here we have to use a variant of Simon's Favourite Factorizing Trick (SFFT). It's just the name of a technique which is used to solve Diophantine Equations.

Simon's Favourite Factorizing Trick:

$(x + 1)(y + 1) = xy + x + y + 1$

It's the same if you put $a = b = 1$ in the following equation:

${(x + a)(y + b) = xy + xb + ay + ab}$

If you just put $a = b = 1$ it becomes SFFT, and it's incredibly useful in solving a lot of Diophantine Equations which has a $"xy"$ term in it.

Now, this SFFT has a lot of variants, and can be used in a number of ways to solve Diophantine Equations.

One of the conclusion/variants is:

$xy + kx + ly = (x + l)(y + k) - kl$

Using that variant in the given question,

$3x + 2y = xy + 1$

$\longrightarrow xy - 3x - 2y = - 1$

$\longrightarrow xy + ( - 3)x + ( - 2)y = - 1$

${\longrightarrow \{x + ( - 2) \} \{y + ( - 3) \} - ( - 3 \times - 2)= - 1}$

$\longrightarrow (x - 2)(y - 3) - 6 = - 1$

$\longrightarrow (x - 2)(y - 3) = 5$

Now, product of two integers is 5, possible pairs include,

• (1, 5)
• (5, 1)
• (-1, 5)
• (1, -5)

So,

First pair:

$x - 2 = 1$

$\longrightarrow x = 3$

And, $y-3 = 5$

$\longrightarrow y = 8$

$\longrightarrow (x,y) = (3,8)$

Second pair:

$x - 2 = 5$

$\longrightarrow x = 7$

And $y - 3 = 1$

$\longrightarrow y = 4$

$\longrightarrow (x,y) = (7,4)$

For third and fourth pairs, on solving them for $x$ and $y$ we find y to be negative in the third pair, and x to be negative in the fourth pair.

But since accroding to the question, $x$ and $y$ are positive integers, we will not consider them.

The possible pairs of $(x,y)$ satisfying $3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+$ are

$\longrightarrow \underline{\underline{(3,8),(7,4)}}$

Answer 2

Step-by-step explanation:

Note: Here \alphaα is denoted by xx and \betaβ is denoted by y.y. ]

\begin{gathered}\\\end{gathered}

Complete question:

3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+.3x+2y=xy+1∀x,y∈Z

+

.

\begin{gathered}\\\end{gathered}

Here we have to use a variant of Simon's Favourite Factorizing Trick (SFFT). It's just the name of a technique which is used to solve Diophantine Equations.

\begin{gathered}\\\end{gathered}

Simon's Favourite Factorizing Trick:

(x + 1)(y + 1) = xy + x + y + 1(x+1)(y+1)=xy+x+y+1

It's the same if you put a = b = 1a=b=1 in the following equation:

{(x + a)(y + b) = xy + xb + ay + ab}(x+a)(y+b)=xy+xb+ay+ab

If you just put a = b = 1a=b=1 it becomes SFFT, and it's incredibly useful in solving a lot of Diophantine Equations which has a "xy""xy" term in it.

\begin{gathered}\\\end{gathered}

Now, this SFFT has a lot of variants, and can be used in a number of ways to solve Diophantine Equations.

One of the conclusion/variants is:

xy + kx + ly = (x + l)(y + k) - klxy+kx+ly=(x+l)(y+k)−kl

\begin{gathered}\\\end{gathered}

Using that variant in the given question,

3x + 2y = xy + 13x+2y=xy+1

\longrightarrow xy - 3x - 2y = - 1⟶xy−3x−2y=−1

\longrightarrow xy + ( - 3)x + ( - 2)y = - 1⟶xy+(−3)x+(−2)y=−1

{\longrightarrow \{x + ( - 2) \} \{y + ( - 3) \} - ( - 3 \times - 2)= - 1}⟶{x+(−2)}{y+(−3)}−(−3×−2)=−1

\longrightarrow (x - 2)(y - 3) - 6 = - 1⟶(x−2)(y−3)−6=−1

\longrightarrow (x - 2)(y - 3) = 5⟶(x−2)(y−3)=5

Now, product of two integers is 5, possible pairs include,

(1, 5)

(5, 1)

(-1, 5)

(1, -5)

So,

First pair:

x - 2 = 1x−2=1

\longrightarrow x = 3⟶x=3

And, y-3 = 5y−3=5

\longrightarrow y = 8⟶y=8

\longrightarrow (x,y) = (3,8)⟶(x,y)=(3,8)

Second pair:

x - 2 = 5x−2=5

\longrightarrow x = 7⟶x=7

And y - 3 = 1y−3=1

\longrightarrow y = 4⟶y=4

\longrightarrow (x,y) = (7,4)⟶(x,y)=(7,4)

For third and fourth pairs, on solving them for xx and yy we find y to be negative in the third pair, and x to be negative in the fourth pair.

But since accroding to the question, xx and yy are positive integers, we will not consider them.

\begin{gathered}\\\end{gathered}

The possible pairs of (x,y)(x,y) satisfying 3x + 2y = xy + 1\:\:\:\:\forall\:\: \:\:x,y\in\mathbb{Z}^+3x+2y=xy+1∀x,y∈Z

+

are

\longrightarrow \underline{\underline{(3,8),(7,4)}}⟶

(3,8),(7,4)