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 March 3rd, 2012, 02:16 PM #1 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 A Fermat-esque challenge A little background for those of you who don't know: Fermat's Last Theorem (which has been proven, though the proof is extremely long and complicated) says that there are no integer solutions to the equation $a^n + b^n= c^n$ if $n \geq 3$. The challenge: Give an elementary proof that there are no integer solutions to the equation $a^n + b^n= c^n$ if $n \geq 3$ and $n+1$ is prime.
 March 7th, 2012, 11:42 AM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: A Fermat-esque challenge A brief summary of my work thus far on the problem: Consider the equation $a^n + b^n= c^n$ mod p. It is clear that each of the numbers $a^n, b^n, c^n$ is congruent to either 0 or 1. The only possible combinations $(a^n, b^n, c^n)$ that will yield solutions are $(0, 0, 0), (1, 0, 1), (0, 1, 1)$. However, if the equation is of the form $(0, 0, 0)$, then we may divide each of the numbers by $p^n$ to produce an equivalent equation and continue the process until at least one of the numbers has no multiples of p left. Thus, we may focus our attention on the case $(1, 0, 1)$, since the other remaining case is identical without loss of generality. Thus, we may write $b^n= y^np^{nk}$ for some integer k such that y is not divisible by p, and the equation becomes (1) $a^n + y^np^{nk}= c^n$. Observing that either exactly one of $a, y, c$ is even or all of them are even, we may focus our attention on cases such that exactly one is even, since if all of them are even, we may divide the equation by a factor of $2^n$ to produce an equivalent equation and continue the process until at least one of the numbers is odd. We also notice that the equation is equivalent to a Pythagorean triple since n is even. Since all primitive Pythagorean triples may be written as $(v^2 - u^2, 2uv, v^2 + u^2)$, where $v > u$, and u and v are relatively prime and of opposite parity, we note that the equation must be a multiple of a Pythagorean triple. It cannot be an even multiple, because then all of the numbers would be even and we have eliminated this case. Therefore, the hypotenuse must be odd; therefore c is odd.
 March 8th, 2012, 05:20 PM #3 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: A Fermat-esque challenge Some additional directions to take: Since n is even, we may write $y^np^{nk}= c^n - a^n = (c^{n/2} - a^{n/2})(c^{n/2} + a^{n/2})$. $p^{nk}= \frac{(c^{n/2} - a^{n/2})(c^{n/2} + a^{n/2})}{y^n}$. $\frac{p^{nk}}{c^{n/2} - a^{n/2}}= \frac{c^{n/2} + a^{n/2}}{y^n}$ and $\frac{p^{nk}}{c^{n/2} + a^{n/2}}= \frac{c^{n/2} - a^{n/2}}{y^n}$ $\frac{p^{nk}(c^{n/2} + a^{n/2})}{c^{n/2} - a^{n/2}}= \frac{(c^{n/2} + a^{n/2})^2}{y^n}$ $\frac{p^{nk}(c^{n/2} - a^{n/2})}{c^{n/2} + a^{n/2}}= \frac{(c^{n/2} - a^{n/2})^2}{y^n}$
 March 9th, 2012, 03:18 PM #4 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: A Fermat-esque challenge I actually discovered a new avenue for the case $a^3 + b^3= c^3$ which I had not previously considered. If you look at the equation mod 7, and observe that $n^3$ mod 7 is either -1, 0, or 1, then you come to the conclusion that either (1) all of the numbers $a, b, c$ are divisible by 7; (2) c is divisible by 7 but a and b are not; (3) c is not divisible by 7 but exactly one of a and b is. In the first case, we may reduce the equation by dividing it by $7^3$ repeatedly until at least one of the numbers is not divisible by 7. In the third case, we will consider the equation (1) $a^3 + 7^{3k}b^3= c^3$, where b is not divisible by 7. We may rearrange the equation to show $b^3= \frac{(c - a)(c^2 + ac + a^2)}{7^{3k}}$. Now, 7 cannot divide both $c - a$ and $c^2 + ac + a^2$, because then c would be congruent to a, and the second equation would become $c^2 + c^2 + c^2= 0$ mod 7, or $3c^2= 0$ mod 7, which is absurd. Therefore, $7^{3k}$ divides one of $c - a$ and $c^2 + ac + a^2$. If it divides $c - a$, we have that $\frac{b^3}{c^2 + ac + a^2}$ is an integer. If it divides $c^2 + ac + a^2$, we have that $\frac{b^3}{c - a}$ is an integer. Assuming the latter, then $b^3= m(c - a)$ for some integer m. Hence equation (1) may be written: $a^3 + 7^{3k}m(c - a)= c^3$ $a^3 - 7^{3k}ma= c^3 - 7^{3k}mc$. Now let us consider the function $f(x)= x^3 - 7^{3k}mx$, which equals 0 at 0 and when $x^2 - 7^{3k}m$, or $x= \sqrt{7^{3k}m}$. The function is decreasing for positive values of x until $f'(x) = 3x^2 - 7^{3k}m = 0$, or $x= \sqrt{\frac{7^{3k}m}{3}}$. Then it increases to infinity. Thus, in order to find 2 different values of x (a and c) such that $f(a)= f(c), c > a=$ , it must be the case that $\sqrt{\frac{7^{3k}m}{3}} < c < \sqrt{7^{3k}m}$, and $0 < a < \sqrt{\frac{7^{3k}m}{3}}$.
 March 12th, 2012, 09:57 AM #5 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: A Fermat-esque challenge If you can prove that no Pythagorean triple consists of three nth powers, then you will have shown that $a^{2n} + b^{2n}= c^{2n}$ has no solutions in integers, since we have $(a^n)^2 + (b^n)^2= (c^n)^2$, which means $(a^n, b^n, c^n)$ must be a Pythagorean triple.

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