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**Peter Norvig** 22 October 2015, revised 28 October 2015
\n", "\n", "# Beal's Conjecture Revisited2\n", "\n", "In 1637, Pierre de Fermat wrote in the margin of a book that he had a proof of his famous \"[Last Theorem](https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem)\":\n", "\n", "> If \$A^n + B^n = C^n\$,\n", ">
where \$A, B, C, n\$ are positive integers\n", ">
then \$n \\le 2\$.\n", "\n", "Centuries passed before [Andrew Beal](https://en.wikipedia.org/wiki/Andrew_Beal), a businessman and amateur mathematician,\n", "made his conjecture in 1993:\n", "\n", "> If \$A^x + B^y = C^z\$, \n", ">
where \$A, B, C, x, y, z\$ are positive integers and \$x, y, z\$ are all greater than \$2\$, \n", ">
then \$A, B\$ and \$C\$ must have a common prime factor.\n", "\n", "[Andrew Wiles](https://en.wikipedia.org/wiki/Andrew_Wiles) proved Fermat's theorem in 1995, but Beal's offer of [\\\$1,000,000](http://abcnews.go.com/blogs/headlines/2013/06/billionaire-offers-1-million-to-solve-math-problem/) for a proof or disproof of *his* conjecture remains unclaimed. I don't have the mathematical skills of Wiles, so all I can do is write a program to search for counterexamples. I first wrote [that program in 2000](http://norvig.com/beal2000.html), and [my name got associated](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=beal%20conjecture) with Beal's Conjecture, which means I get a lot of emails with purported proofs or counterexamples (many asking how they can collect their prize money). So far, all the emails have been wrong. This page catalogs some of the more common errors—including two mistakes of my own—and shows an updated program." ] }, { "cell_type": "markdown", "metadata": { "button": false, "collapsed": false, "deletable": true, "new_sheet": false, "run_control": { "read_only": false } }, "source": [ "# How to Not Win A Million Dollars\n", "\n", "\n", "* A proof must show that there are no examples that satisfy the conditions. A common error is to show how a certain technique generates an infinite collection of numbers that satisfy \$A^x + B^y = C^z\$ and then show that in all of these, \$A, B, C\$ have a common factor. But that's not good enough, unless you can also prove that no other technique can satisfy the equation.\n", "\n", "

\n", "\n", "* It is valid to use proof by contradiction: assume the conjecture is true, and show that that leads to a contradiction. It is not valid to use proof by circular reasoning: assume the conjecture is true, put in some irrelevant steps, and show that it follows that the conjecture is true.\n", "\n", "\n", "

\n", "\n", "* A valid counterexample needs to satisfy all four conditions—don't leave one out:\n", "\n", "> \$A, B, C, x, y, z\$ are positive integers
\n", "\$x, y, z > 2\$
\n", "\$A^x + B^y = C^z\$
\n", "\$A, B, C\$ have no common prime factor.\n", "\n", "(If you think you have a valid counterexample, before you share it with Andrew Beal, or me, or anyone else, you can check it with my [Online Beal Counterexample Checker](http://norvig.com/bealcheck.html).)\n", "\n", "

\n", "\n", "* One correspondent claimed that \$27^4 + 162 ^ 3 = 9 ^ 7\$ was a solution, because the first three conditions hold, and the common factor is 9, which isn't a prime. But of course, if \$A, B, C\$ have 9 as a common factor, then they also have 3, and 3 is prime. The phrase \"no common prime factor\" means the same thing as \"no common factor greater than 1.\"\n", "\n", "

\n", "\n", "* Another claimed that \$2^3+2^3=2^4\$ was a counterexample, because all the bases are 2, which is prime, and prime numbers have no prime factors. But that's not true; a prime number has itself as a factor.\n", "\n", "

\n", "\n", "* A creative person offered \$1359072^4 - 940896^4 = 137998080^3\$, which fails both because \$3^3 2^5 11^2\$ is a common factor, and because it has a subtraction rather than an addition (although, as Julius Jacobsen pointed out, that can \n", "\n", "

\n", "\n", "* Another Beal fan started by saying \"Let \$C = 43\$ and \$z = 3\$. Since \$43 = 21 + 22\$, we have \$43^3 = (21^3 + 22^3).\$\" But of course \$(a + b)^3 \\ne (a^3 + b^3)\$. This fallacy is called [the freshman's dream](https://en.wikipedia.org/wiki/Freshman%27s_dream) (although I remember having different dreams as a freshman).\n", "\n", "