●need detailed solution!
Answers
[Note: Here is denoted by and is denoted by ]
Complete question:
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:
It's the same if you put in the following equation:
If you just put it becomes SFFT, and it's incredibly useful in solving a lot of Diophantine Equations which has a 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:
Using that variant in the given question,
Now, product of two integers is 5, possible pairs include,
- (1, 5)
- (5, 1)
- (-1, 5)
- (1, -5)
So,
First pair:
And,
Second pair:
And
For third and fourth pairs, on solving them for and 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, and are positive integers, we will not consider them.
The possible pairs of satisfying are
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)