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Does Fermat's Last Theorem or Euler's proof for the case n = 3 imply that $x*a^3+y*b^3 = c^3$ has no solutions in integers for $a,b,c$ positive integers and integers $x$ and $y$?

My question comes from me trying to provide an elementary proof of Beal's Conjecture $(A^x + B^y = C^z$ iff $gcd(A,B,C) > 1$ for positive integers $A,B,C,x,y,z$ and $x,y,z >2$). My attack was to assume that $gcd(A,B,C) = 1$ which implied that A,B,C were pairwise mutually coprime. Otherwise, they all share a prime factor (easy to show). Thus, $gcd(A,B) = 1$ implies $gcd(A^x,B^y) = 1$. Therefore, by Bezout's Identity there exists integers $z_1$ and $z_2$ such that,

(1) $$A^x*z_1 + B^y*z_2 =1.$$

Since $x,y > 2$, we have $x= 3 + s$ and $y =3 + r$. Thus, let $m = C^3*A^s$ and $n = C^3*A^r$, then (1) becomes,

(2) $$m*A^3 + n*B^3 = C^3.$$

if (2) has no solutions in positive coprime integers A,B and integers m and n, then there would be a contradiciton, and $gcd(A,B) > 1$ implying $gcd(A,B,C) >1$ and Beal's conjecture would follow. However, you all and others in private emails have provided a way to construct infinitely many counterexamples to my reasoning. Can this line of reasoning be fixed? Maybe an examination of Elliptic Curves and Modular Forms, etc. Thank you.

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  • $\begingroup$ I should add the condition that a,b,c,x,y cannot be 1 and the gcd(a,b) = 1. Can anyone find a counterexample with this added condition? $\endgroup$ – Avery Carr Oct 5 '15 at 16:57
  • $\begingroup$ see my edited answer. $\endgroup$ – Aloizio Macedo Oct 5 '15 at 16:59
  • $\begingroup$ $19\cdot 1^3+1\cdot 2^3=3^3$. Multiple infinity of others. Too many variables, too little constraint. $\endgroup$ – André Nicolas Oct 5 '15 at 17:17
  • $\begingroup$ There is no point in restricting to $a,b,c$ positive, since you can absorb the signs into $x,y$. $\endgroup$ – user14972 Oct 5 '15 at 17:18
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    $\begingroup$ No minor tinkering with the conditions on $a,b,c,x,y$ will yield "no non-trivial solution." $\endgroup$ – André Nicolas Oct 5 '15 at 20:35
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$2\cdot 2^3-1\cdot2^3=2^3$

And even if $x,y$ must be positive:

$2^2\cdot2^3+2^2\cdot2^3=4^3.$

And to your last request...

$24\cdot3^3-80\cdot2^3=2^3.$

Please, check if those are the real conditions you need.

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  • $\begingroup$ Thank you for your quick comment. Can you find a counterexample to with the added conditions? $\endgroup$ – Avery Carr Oct 5 '15 at 17:01
  • $\begingroup$ I edited my answer accordingly. Please check if are the definite conditions you have, and refrain from "updating" questions. $\endgroup$ – Aloizio Macedo Oct 5 '15 at 17:17
  • $\begingroup$ Thank you Aloizio, Those are the conditions met. I will refrain from updating questions in the future. $\endgroup$ – Avery Carr Oct 5 '15 at 22:04
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Absolutely not!

Example. If a and b are relatively prime you can fin xa + yb = 1. (This is Euclid's algorithm.) So $a^n* and *b^n$ are also relatively prime so you can fine $x*a^n + y*b^n = 1$ so you can find $x*a^n + y*b^n = c^n$

Try 2, 3, and 5. $2^3 = 8$ and $3^3 = 27$ so we can find $8x + 27y = 1$ e.g. x = -10 and y = 3. So $-1250*2^3 + 450*3^3 = 5^3$.

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For everything strictly positive, let $a=b$ and write

$$(x+y) = \left(\frac{c}{a} \right)^3$$

which obviously has solutions. For instance, take $c=4, a=2$, $x=2$, $y=6$.

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  • $\begingroup$ Thank you so much for your prompt comment. Can you find a counterexample with the above added conditions? $\endgroup$ – Avery Carr Oct 5 '15 at 17:02
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All pairwise coprime $a,b,c,x,y$ with $21\ge a>b\ge 2$ and $x,y\in[2,100]$:

$$\begin{align} 55\cdot 3^3 + 89\cdot 2^3 &= 13^3 \\ 71\cdot 3^3 + 35\cdot 2^3 &= 13^3 \\ 7\cdot 5^3 + 57\cdot 2^3 &= 11^3 \\ 17\cdot 5^3 + 9\cdot 2^3 &= 13^3 \\ 73\cdot 5^3 + 17\cdot 2^3 &= 21^3 \\ 91\cdot 5^3 + 99\cdot 2^3 &= 23^3 \\ 17\cdot 5^3 + 73\cdot 3^3 &= 16^3 \\ 19\cdot 5^3 + 94\cdot 3^3 &= 17^3 \\ 82\cdot 5^3 + 71\cdot 3^3 &= 23^3 \\ 89\cdot 7^3 + 83\cdot 3^3 &= 32^3 \\ 37\cdot 14^3 + 85\cdot 3^3 &= 47^3 \\ 73\cdot 16^3 + 65\cdot 3^3 &= 67^3 \\ 53\cdot 17^3 + 65\cdot 3^3 &= 64^3 \\ 39\cdot 5^3 + 31\cdot 4^3 &= 19^3 \\ 17\cdot 7^3 + 99\cdot 4^3 &= 23^3 \\ 29\cdot 13^3 + 83\cdot 5^3 &= 42^3 \\ 66\cdot 13^3 + 31\cdot 5^3 &= 53^3 \\ 17\cdot 16^3 + 79\cdot 5^3 &= 43^3 \\ 3\cdot 19^3 + 11\cdot 5^3 &= 28^3 \\ 34\cdot 11^3 + 69\cdot 7^3 &= 41^3 \\ 43\cdot 13^3 + 47\cdot 7^3 &= 48^3 \\ 71\cdot 16^3 + 29\cdot 7^3 &=67^3 \\ 53\cdot 11^3 + 65\cdot 8^3 &= 47^3 \\ 19\cdot 10^3 + 83\cdot 9^3 &= 43^3 \\ 26\cdot 17^3 + 53\cdot 9^3 &= 55^3 \\ 85\cdot 13^3 + 14\cdot 11^3 &= 59^3 \\ 2\cdot 17^3 + 15\cdot 11^3 &= 31^3 \end{align}$$

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Chose two arbitrary cubes $c^3, b^3$ and form the set $A$={$c^3-yb^3$ |$y\in \mathbb Z$}. If all element of $A$ is cube-free you have discovered a remarkable couple of cubes $(c^3,b^3)$. Take then $N\in\mathbb Z$ such that $N=xa^3$. You can get this way infinitely many $(x,y)$ satisfying your equation.

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