- #1

- 64

- 0

## Main Question or Discussion Point

Sorry guys about yesterday. I did a fundamental mistake in proof and I was so embaressed, that I have decided, I will give it a try for the last time. Here is the proof

Proof:

Let us suppose that a,b,c are coprimes, so if we construct the from a,b,c the smallest triangle for solution of the Fermats Last Theorem.

so lets suppose that the sollution exist, a^n + b^n = c^n lets suppose a,b,c are coprimes

Let us check the problem for odd powers of n. We can write now the equation (c^n + b^n)*(c^n - b^n) = (c^2n - b^2n) so that holds everytime, not specifically for the problem. ---------------> now the Fermats Last theorem is included: if c^n - b^n = a^n than we can easily see that the factors on the left are coprimes. (c^n + b^n) equals 2b^n + a^n and that shure is coprime to a, since a and b are coprimes. So lets rearange the equation for odd powers of n. (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c^2 - b^2)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (c - b)*(c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> so we can exclude (c - b) factor from the eqation: (c^n + b^n)*((c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-------------------------------------->so since we can rearange the expresion: c^(n-1) + c^(n-2)*b..............+ b^(n-1) into c^(n-1) + c^(n-2)*b = c^(n-2)*(c+b) first two are devidable by (c+ b) we carry on doing that and we see if we take first two members and than second two and third two, we see that all are devidable by (b + c) BUT since n is odd we have the odd numbers of those members so we come till the last one: and we get somehow: (b + c)*z + b^(n-1) so (c + b) and (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) are coprimes. So the only possibility is now, that c^n + b^n is devidable by (c + b) if the solution exists. Ok lets suppose that c^n + b^n = (c + b)*l where l is the member of natural numbers ---------------->c*l - c^n = b^n -l*b ------------------------>c*(l - c^(n-1) = b*(b^(n-1) -l) -----------------> (l+c^(n-1)) = b*m and (b^(n-1)-l) = c*m since b and c are coprimes.now sum both equations: (l+c^(n-1)) + (b^(n-1)-l) = b*m + c*m, l goes out so: c^(n-1) + (b^(n-1) = m* (c + b) where m is the member of natural numbers. So if we state that c^n + b^n is devidable by (b + c) it follows out of that that c^(n-1) + b^(n-1) is devidable by (b + c). Since (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) has odd number of members when n is odd, we saw that (c+ b ) does not devide that expression, now if we start from the begining and sum 2 by 2 till the last one is out: (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (b + c) *z + b^(n-1), now lets turn this procedure around: lets sum all the members till the last one 2 by 2 by the backside so: c*b^(n-2) + b^(n-1) = b^(n-2)*(c + b)..........so we carry on till we come to the c^(n-1) -------------> c^(n-1) + (c + b)*t where t is again number of natural numbers. BUT if we multiply (c^(n-1) + c^(n-2)*b..............+ b^(n-1) by 2 we can sum the equations: c^(n-1) + (c + b)*t + (b + c) *z + b^(n-1) since (b + c) devides c^(n-1) + b^(n-1), we got that if we multiply the expression by 2 (c+ b ) devides 2*(c^(n-1) + c^(n-2)*b..............+ b^(n-1)) since (c + b) is odd -------------------------------> (c+b) devides (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) too. Thats a contradiction (c + b) is coprime to (c^(n-1) + c^(n-2)*b..............+ b^(n-1), so (c^n + b^n) and(c^n - b^n) are coprimes when a,b,c are coprimes -------------------> so (c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1)) is the member of rational numbers. This is again a contradiction, since we know that this expression is the member of natural numbers. So we can not find solutions in whole number sistem of a,b,c for n is odd and > 2.

I want to make better impression, so I got back to the problem for a few hours today.

Sincerelly

Robert Kulovec Mueller, Slovenija

Proof:

Let us suppose that a,b,c are coprimes, so if we construct the from a,b,c the smallest triangle for solution of the Fermats Last Theorem.

so lets suppose that the sollution exist, a^n + b^n = c^n lets suppose a,b,c are coprimes

Let us check the problem for odd powers of n. We can write now the equation (c^n + b^n)*(c^n - b^n) = (c^2n - b^2n) so that holds everytime, not specifically for the problem. ---------------> now the Fermats Last theorem is included: if c^n - b^n = a^n than we can easily see that the factors on the left are coprimes. (c^n + b^n) equals 2b^n + a^n and that shure is coprime to a, since a and b are coprimes. So lets rearange the equation for odd powers of n. (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c^2 - b^2)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> (c^n + b^n)*(c - b)* (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (c - b)*(c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-----------> so we can exclude (c - b) factor from the eqation: (c^n + b^n)*((c^(n-1) + c^(n-2)*b..............+ b^(n-1) = (c + b)*(c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1))-------------------------------------->so since we can rearange the expresion: c^(n-1) + c^(n-2)*b..............+ b^(n-1) into c^(n-1) + c^(n-2)*b = c^(n-2)*(c+b) first two are devidable by (c+ b) we carry on doing that and we see if we take first two members and than second two and third two, we see that all are devidable by (b + c) BUT since n is odd we have the odd numbers of those members so we come till the last one: and we get somehow: (b + c)*z + b^(n-1) so (c + b) and (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) are coprimes. So the only possibility is now, that c^n + b^n is devidable by (c + b) if the solution exists. Ok lets suppose that c^n + b^n = (c + b)*l where l is the member of natural numbers ---------------->c*l - c^n = b^n -l*b ------------------------>c*(l - c^(n-1) = b*(b^(n-1) -l) -----------------> (l+c^(n-1)) = b*m and (b^(n-1)-l) = c*m since b and c are coprimes.now sum both equations: (l+c^(n-1)) + (b^(n-1)-l) = b*m + c*m, l goes out so: c^(n-1) + (b^(n-1) = m* (c + b) where m is the member of natural numbers. So if we state that c^n + b^n is devidable by (b + c) it follows out of that that c^(n-1) + b^(n-1) is devidable by (b + c). Since (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) has odd number of members when n is odd, we saw that (c+ b ) does not devide that expression, now if we start from the begining and sum 2 by 2 till the last one is out: (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) = (b + c) *z + b^(n-1), now lets turn this procedure around: lets sum all the members till the last one 2 by 2 by the backside so: c*b^(n-2) + b^(n-1) = b^(n-2)*(c + b)..........so we carry on till we come to the c^(n-1) -------------> c^(n-1) + (c + b)*t where t is again number of natural numbers. BUT if we multiply (c^(n-1) + c^(n-2)*b..............+ b^(n-1) by 2 we can sum the equations: c^(n-1) + (c + b)*t + (b + c) *z + b^(n-1) since (b + c) devides c^(n-1) + b^(n-1), we got that if we multiply the expression by 2 (c+ b ) devides 2*(c^(n-1) + c^(n-2)*b..............+ b^(n-1)) since (c + b) is odd -------------------------------> (c+b) devides (c^(n-1) + c^(n-2)*b..............+ b^(n-1)) too. Thats a contradiction (c + b) is coprime to (c^(n-1) + c^(n-2)*b..............+ b^(n-1), so (c^n + b^n) and(c^n - b^n) are coprimes when a,b,c are coprimes -------------------> so (c^2(n-1) + c^2(n-2)*b^2 ..............+ b^2(n-1)) is the member of rational numbers. This is again a contradiction, since we know that this expression is the member of natural numbers. So we can not find solutions in whole number sistem of a,b,c for n is odd and > 2.

I want to make better impression, so I got back to the problem for a few hours today.

Sincerelly

Robert Kulovec Mueller, Slovenija