ფაილების კატალოგი

Discovery #298
Conjecture (Papava). If gcd(a, b)=1 and a+b=c then
log[rad(a), a] log[rad(b), b] log[rad(c), c]≤rad(abc).

Discovery #297
Conjecture (Papava). If gcd(a,b)=1 and a+b=c then
c<rad(abc)log[rad(a), a]log[rad(b), b]log[rad(c), c].

Result 1. Fermat's Last Theorem.
An equation x^n+y^n=z^n, n>2 has not a solution.
Proof:
z^n<xyz log₂x^n log₂y^n log₂z^n<z³n³(log₂z)³<z³z³z³=z⁹
n<9.

Result 2. Beal conjecture (Prize $1000 000).
...

Papava's Last Conjecture

There is a constant μ>1 such that for each ε>0 there are only finitely many triples of coprime positive integers a, b, c for which
a+b=c and c>(gpf(a)gpf(b)gpf(c))^(μ+ε).

Definition. For an integer n≥2, let gpf(n) denote the greatest prime factor of n.

Conjecture (Papava). n<gpf(n⁹-1).

A discovery #294 (Papava)

Definition. For an integer n≥2, let gpf(n) denote the greatest prime factor of n.

Conjecture. If a,b,c∈N, gcd(a,b)=1 and a+b=c then c<(gpf(abc))⁹.

Result 1. Fermat's Last Theorem;
Result 2. Beal conjecture (Prize $1000 000);
...

THE MOST BEAUTIFUL AND STRONGEST PAPAVA'S CONJECTURE IN THE XXI CENTURY.

Definition. For an integer n≥2, let gpf(n) denote the greatest prime factor of n.

Conjecture (Papava). If a,b,c∈N, a+b=c, gcd(a,b)=1 then
c<(gpf(a)gpf(b)gpf(c))³.

Result 1. Fermat's Last Theorem;
Result 2. Beal conjecture (Prize $1000 000);
Result 3. ABC conjecture when ε=2;
...

...

...

a≠b

...