explain fermats last theorem full explanation and formula explanation who will give ans I will Mark as brainlist but give correct answer
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mah21
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c satisfy the equation an + bn = cn for any integer value of n greater than 2. The cases n = 1 and n = 2 have been known to have infinitely many solutions since antiquity
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mah21
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mahisharma21
Fermat's Last Theorem simple proofup vote1down votefavorite8I came across this simple proof of Fermat's last theorem. Some think it's legit. Some argued that the author's assumptions are flawed. It's rather lengthy but the first part goes like this:Let x,yx,y be 22 positive non-zero coprime integers and nn an integer greater than 22. According to the binomial theorem:(x+y)n=∑k=0n(nk)xn−kyk(x+y)n=∑k=0n(nk)xn−kykthen,(x+y)n−xn=nxn−1y+∑k=2n−1(nk)xn−kyk+yn(x+y)n−xn=nxn−1y+∑k=2n−1(nk)xn−kyk+yn(x+y)n−xn=y(nxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1)(x+y)n−xn=y(nxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1)y(nxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1)=zny(nxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1)=znIn the first case, he assumed that the 2 factors are coprime when gcd(y,n)=1gcd(y,n)=1 . Then he wrote:y=qny=qnnxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1=pnnxn−1+∑k=2n−1(nk)xn−kyk−1+yn−1=pnBy replacing yy by qnqn,nxn−1+∑k=2n−1(nk)xn−kqn(k−1)+qn(n−1)=pn(∗)nxn−1+∑k=2n−1(nk)xn−kqn(k−1)+qn(n−1)=pn(∗)from this bivariate polynomial,he fixed alternatively xx and y=qny=qn and by applying the rational root theorem, he obtainedqn(n−1)−pn=nxtqn(n−1)−pn=nxtandnxn−1−pn=qnsnxn−1−pn=qns(s,ts,t non-zero integers) by equating pxpx:qn(n−1)−sqn=nx(t−xn−2)qn(n−1)−sqn=nx(t−xn−2)Then, he uses one of the trivial solutions of Fermat's equations. He wrote, when x+y=1x+y=1,if x=0x=0then y=1y=1 and vice versa.Therefore, he wrote: x=0x=0 iff qn(n−1)=sqnqn(n−1)=sqn, he obtains:q=1q=1ors=qn−2s=qn−2By substituting ss by qn−2qn−2 in nxn−1−pn=qnsnxn−1−pn=qns, he obtains:nxn−1−pn=qn(n−1)nxn−1−pn=qn(n−1)Then, he replace that expression in equation (*) and pointed out that:∑k=2n−1(nk)xn−kqn(k−1)=0∑k=2n−1(nk)xn−kqn(k−1)=0. Since x,y=qnx,y=qn are positive integers for all n>2n>2, a sum of positive numbers can not be equal to zero. Which leads to a contradiction.
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