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Beal.ipynb

@@ -76,6 +76,10 @@
     "\n",
     "<p>\n",
     "\n",
+    "* Mustafa Pehlivan came up with an example involving 76-million-digit numbers, which took some work to prove wrong (by 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",
     "\n",
     "<p>\n",
@@ -184,7 +188,6 @@
    "source": [
     "They say \"close\" only counts in horseshoes and hand grenades, and if you threw two horseshoes at a stake on the planet [Kapteyn-b](https://en.wikipedia.org/wiki/Kapteyn_b) (a possibly habitable and thus possibly horseshoe-playing exoplanet 12.8 light years from Earth) and the two paths differed in the 19th digit, the horseshoes would end up [less than an inch](https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=12.8%20light%20years%20*%201e-19%20in%20inches) apart. That's really, really close, but close doesn't count in number theory.\n",
     "\n",
-    "Another near-solution was proposed by Mustafa Pehlivan: $193^{33554432} + 33554432^{193} = 37249^{16777216}$. Both $193^{33554432}$ and $37249^{16777216}$ are 76,690,578-digit numbers (while $33554432^{193}$ is a puny 1,453 digits), so it is possible that the two sides are equal, and in fact they agree in all but the last 4 of the 76 million digits. There is no analogy I can make in the physical universe that is that close.\n",
     "\n",
     "# *The Simpsons* and Fermat\n",
     "\n",
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@@ -679,6 +684,7 @@
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