Linear Diophantine equation of different form

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

Taking the equation axy+bx+cy+d=0axy+bx+cy+d=0 xy+ba+ca+da=0xy+\frac{b}{a}+\frac{c}{a}+\frac{d}{a}=0 x(y+ba)+ca(y+ba)cba2+da=0x(y+\frac{b}{a})+\frac{c}{a}(y+\frac{b}{a})-\frac{cb}{a^2}+\frac{d}{a}=0 (x+ca)(y+ba)=cba2da(x+\frac{c}{a})(y+\frac{b}{a})=\frac{cb}{a^2}-\frac{d}{a} (x+ca)(y+ba)=cba2da(x+\frac{c}{a})(y+\frac{b}{a})=\frac{cb}{a^2}-\frac{d}{a} Now as aa divides bb and cc both a2bc\therefore a^2|bc cba2Z\therefore \frac{cb}{a^2}∈Z Also as ada|d daZ\therefore \frac{d}{a}∈Z cba2daZ,cba2da=k\therefore \frac{cb}{a^2}-\frac{d}{a}∈Z , \frac{cb}{a^2}-\frac{d}{a}=k (x+ca)(y+ba)=k(x+\frac{c}{a})(y+\frac{b}{a})=k Now finding factors of kk will give you possible pair of integer values of xx and yy


  • 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=0axy+bx+cy+d=0) then please comment.

  • aba|b denotes that "aa divides bb"

Note by Zakir Husain
3 weeks, 6 days ago

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