# Linear Diophantine equation of different form

Most of you shall be familiar to linear Diophantine equations of the form $ax+by+c=0$, but in this discussion I will talk about linear Diophantine of the form $axy+bx+cy+d=0$ where $a|b,a|c,a|d$ and $a,b,c$ and $d$ are integers.

Taking the equation $axy+bx+cy+d=0$ $xy+\frac{b}{a}+\frac{c}{a}+\frac{d}{a}=0$ $x(y+\frac{b}{a})+\frac{c}{a}(y+\frac{b}{a})-\frac{cb}{a^2}+\frac{d}{a}=0$ $(x+\frac{c}{a})(y+\frac{b}{a})=\frac{cb}{a^2}-\frac{d}{a}$ $(x+\frac{c}{a})(y+\frac{b}{a})=\frac{cb}{a^2}-\frac{d}{a}$ Now as $a$ divides $b$ and $c$ both $\therefore a^2|bc$ $\therefore \frac{cb}{a^2}∈Z$ Also as $a|d$ $\therefore \frac{d}{a}∈Z$ $\therefore \frac{cb}{a^2}-\frac{d}{a}∈Z , \frac{cb}{a^2}-\frac{d}{a}=k$ $(x+\frac{c}{a})(y+\frac{b}{a})=k$ Now finding factors of $k$ will give you possible pair of integer values of $x$ and $y$

Note:

• I have done this for a particular case of the equation if you find any other result related to the equation ($axy+bx+cy+d=0$) then please comment.

• $a|b$ denotes that "$a$ divides $b$"

Note by Zakir Husain
1 year, 1 month ago

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