setting and and b |=0 the equation becomes 11a+11b=ab ab-11a-11b+121-121=0 a(b-11)-11(b-11)-121=0 (a-11)(b-11)=121 Because of comutativity any solution pair (m,n) will also include (n,m) but to avoid repetitions i will write only one set of solutions for pairs of factors of 121: (-1)*(-121),1*121,(-11)*(-11),11*11 { (10,-110), (12,132),(0,0) not accepted,(22,22)}
so for any integers
1/a+ 1/b =1/c
then a+b =4c or (c+1)^2
Very nice! ❤
setting and and b |=0 the equation becomes
11a+11b=ab
ab-11a-11b+121-121=0
a(b-11)-11(b-11)-121=0
(a-11)(b-11)=121
Because of comutativity any solution pair (m,n) will also include (n,m) but to avoid repetitions i will write only one set of solutions for pairs of factors of 121: (-1)*(-121),1*121,(-11)*(-11),11*11
{ (10,-110), (12,132),(0,0) not accepted,(22,22)}
Very nice! ❤
a、ｂが整数とは書いていない。だから与式を満たす整数でないa,bは無限に存在することになる。
初めに a>0 かつ b>0 と言われています。❤
@@SALogics
a＝11(1+1/√2) , b=11(1＋√2) でも与式は成り立つ。