Fermat’s Last Theorem is a famous theorem in number theory which is difficult to prove.However,it is known that the version of polynomials with one variable of Fermat’s Last Theorem over C can be proved very concisely...Fermat’s Last Theorem is a famous theorem in number theory which is difficult to prove.However,it is known that the version of polynomials with one variable of Fermat’s Last Theorem over C can be proved very concisely.The aim of this paper is to study the similar problems about Fermat’s Last Theorem for multivariate(skew)-polynomials with any characteristic.展开更多
In this paper, some conmments on the proof of Fermat’s last theorem are proposed.The main resuilt is thai the proof proposed by Wong Chiahe is only part of proof for fermat’s last theorem. That is to sqy ,the proof...In this paper, some conmments on the proof of Fermat’s last theorem are proposed.The main resuilt is thai the proof proposed by Wong Chiahe is only part of proof for fermat’s last theorem. That is to sqy ,the proof is not all-full proof to Fermat’s last theorem.展开更多
i) Instead of x￣n+ y￣n = z￣n ,we use as the general equation of Fermat's Last Theorem (FLT),where a and b are two arbitrary natural numbers .By means of binomial expansion ,(0.1) an be written as Because a￣...i) Instead of x￣n+ y￣n = z￣n ,we use as the general equation of Fermat's Last Theorem (FLT),where a and b are two arbitrary natural numbers .By means of binomial expansion ,(0.1) an be written as Because a￣r-(-b)￣r always contains a +b as its factor ,(0.2) can be written as where φ_r =[a￣r-(-b)￣r]/ (a+b ) are integers for r=1 . 2, 3. ...n (ii) Lets be a factor of a+b and let (a +b) = se. We can use x= sy to transform (0.3 ) to the following (0.4)(iii ) Dividing (0.4) by s￣2 we have On the left side of (0.5) there is a polynomial of y with integer coefficient and on the right side there is a constant cφ/s .If cφ/s is not an integer ,then we cannot find an integer y to satisfy (0.5), and then FLT is true for this case. If cφ_n/s is an integer ,we may change a and c such the cφ_n/s≠an integer .展开更多
Let p be a prime with p≡3(mod 4). In this paper,by using some results relate the representation of integers by primitive binary quadratic forms,we prove that if x,y,z are positive integers satisfying x^p+y^p=z^p, p|x...Let p be a prime with p≡3(mod 4). In this paper,by using some results relate the representation of integers by primitive binary quadratic forms,we prove that if x,y,z are positive integers satisfying x^p+y^p=z^p, p|xyz, x<y<z, then y>p^(6p-2)/2.展开更多
基金supported by the National Natural Science Foundation of China(12131015,12071422).
文摘Fermat’s Last Theorem is a famous theorem in number theory which is difficult to prove.However,it is known that the version of polynomials with one variable of Fermat’s Last Theorem over C can be proved very concisely.The aim of this paper is to study the similar problems about Fermat’s Last Theorem for multivariate(skew)-polynomials with any characteristic.
文摘In this paper, some conmments on the proof of Fermat’s last theorem are proposed.The main resuilt is thai the proof proposed by Wong Chiahe is only part of proof for fermat’s last theorem. That is to sqy ,the proof is not all-full proof to Fermat’s last theorem.
文摘i) Instead of x￣n+ y￣n = z￣n ,we use as the general equation of Fermat's Last Theorem (FLT),where a and b are two arbitrary natural numbers .By means of binomial expansion ,(0.1) an be written as Because a￣r-(-b)￣r always contains a +b as its factor ,(0.2) can be written as where φ_r =[a￣r-(-b)￣r]/ (a+b ) are integers for r=1 . 2, 3. ...n (ii) Lets be a factor of a+b and let (a +b) = se. We can use x= sy to transform (0.3 ) to the following (0.4)(iii ) Dividing (0.4) by s￣2 we have On the left side of (0.5) there is a polynomial of y with integer coefficient and on the right side there is a constant cφ/s .If cφ/s is not an integer ,then we cannot find an integer y to satisfy (0.5), and then FLT is true for this case. If cφ_n/s is an integer ,we may change a and c such the cφ_n/s≠an integer .
文摘Let p be a prime with p≡3(mod 4). In this paper,by using some results relate the representation of integers by primitive binary quadratic forms,we prove that if x,y,z are positive integers satisfying x^p+y^p=z^p, p|xyz, x<y<z, then y>p^(6p-2)/2.
