Emzari Papava is answering: Mr. Wythagoras, 1. m>=(rad(m))^(1+ε) true always when ε=0, but when ε>0 then _ no; 2. Induction: first step: if c=rad(c) then abc conjecture true. second step: let abc conjecture true for such m that m<c and from it prov abc conjecture for c; 3. gcd(x,y)=1, x+y=m -->gcd(x,m)=1 and gcd(y,m)=1; 4. You are not even wrong; 5. You are not even wrong; 6. From induction: let n<=(rad(n))^(1+ε) , (rad(m))^(1+ε)<m, gcd(m,n)=1, m=x+y, gcd(x,y)=1 then c=mn<(rad(xym))^(1+ε)*(rad(n))^(1+ε).
@Wythagoras sayng: Okay, I will list the worst, most horrible mistakes here: 1. He claims that m>(rad(m))^(1+ε). rad(m) is the product of distinct prime factors of m, therefore always smaller than or equal to m. And why does he need it? 2. Induction? I don't see base case, induction hypothesis, induction step? 3. rad(xym)rad(n)=rad(xy)rad(mn). What if gcd(x,m)≠1 or gcd(y,m)≠1? 4. His proof can't be right anyway. Because if it was legit, it would show that there are no solutions, which is not true. 5. The rows after it are some gibberish to me, no idea what he wants, because he doesn't give any words to explain his 'proof'. 6. I would be pleased to know why mn<rad(xym)rad(n). Emzari Papava is answering: Mr. Wythagoras, 1. m>=(rad(m))^(1+ε) true always when ε=0, but when ε>0 then _ no; 2. Induction: first step: if c=rad(c) then abc conjecture true. second step: let abc conjecture true for such m that m<c and from it prov abc conjecture for c; 3. gcd(x,y)=1, x+y=m -->gcd(x,m)=1 and gcd(y,m)=1; 4. You are not even wrong; 5. You are not even wrong; 6. From induction: let n<=(rad(n))^(1+ε) , (rad(m))^(1+ε)<m, gcd(m,n)=1, m=x+y, gcd(x,y)=1 then c=mn<(rad(xym))^(1+ε)*(rad(n))^(1+ε).
