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ipynb/Beal.ipynb
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@ -33,44 +33,34 @@
"\n",
"# Online Beal Counterexample Checker\n",
"\n",
"Got a counterexample? Check it with my **[Online Beal Counterexample Checker](http://norvig.com/bealcheck.html)**.\n",
"Got a counterexample? Check it with my [Online Beal Counterexample Checker](http://norvig.com/bealcheck.html).\n",
"\n",
"# 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 pattern generates an infinite number of $(A, x, B, y, C, z)$ examples, and that the conjecture holds for this entire infinite collection. But that's not good enough, unless you can also prove that the conjecture holds for every other possible pattern.\n",
"\n",
"<p>\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",
"<p>\n",
"\n",
"* A valid counterexample needs to satisfy all four conditions&mdash;don't leave one out.\n",
"\n",
"\n",
"<p>\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. \"No common prime factor\" means the same thing as \"no common factor greater than 1.\"\n",
"\n",
"<p>\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",
"<p>\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, it could be rewritten as $137998080^3 + 940896^4 = 1359072^4$).\n",
"\n",
"<p>\n",
"\n",
"* Mustafa Pehlivan came up with an example involving 76-million-digit numbers, which took some work to prove wrong (using modulo arithmetic). \n",
"\n",
"<p>\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",
"* 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",
"<p>\n",
"\n",
"* Multiple people proposed counterexamples similar to this one:"
]