**Uncracked 114**

##### (Diophantus of Alexandria, c. 215-290)

Students working with exponents should be asked to find which values of n from 0-100 are possible by summing the cubes of three * positive* integers: a^3 + b^3 + c^3 = n.

Next, ask if the cubes can be * negative*. This is unsolved as you can see in the video, but do not tell your students this till the end of class! This video might be the best way to end the class – or to start the next class.

http://mathoverflow.net/ is a recommended resource for math entusiasts. It shows some of the history of this problem for discoveries n≤100:

*(1960s)*

*87=4271³–4126³–1972³*

*96=−15250³+13139³+10853³*

*91=83538³–67134³–65453³*

*80=−112969³+103532³+69241³*

*(1990s)*

*39=−159380³+134476³+117367³*

*75–435203231³+435203083³+4381159³*

*84=41639611³–41531726³–8241191³*

*(2000s)*

*30=2220422932³–2218888517³–283059965³*

*52=−61922712865³+23961292454³+60702901317³*

*74=−284650292555885³+66229832190556³+283450105697727³*

*For n≤1000, the problem is still open only for 33, 42, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, and 975 *

This problem also lends itself to algebraic exploration, but only for elite classes. As well as the algebraic exploration seen in the video check out Euler-Binet solutions here.

________________________

Hi Gord,

Just thought you should know significant progress has been made; the website currently reads “For n≤1000, the problem is still open only for 33, 42, 114, 165, 390, 579, 627, 633, 732, 795, 906, 921, and 975”

In 2019, both 33 and 42 have been solved (the former by Andrew Booker, and the latter by Andrew Booker and Andrew Sutherland):

33 = (-8 778 405 442 862 239)^3 + (-2 736 111 468 807 040)^3 + (8 866 128 975 287 528)^3

42 = (-80 538 738 812 075 974)^3 + (80 435 758 145 817 515)^3 + (12 602 123 297 335 631)^3

As far as I’m aware, the other 11 numbers on your list remain the only unsolved cases below 1000.

Cheers,

Vince

Monday, April 4, 2022

________________________

I was wrong – there are other solutions that Booker found in 2019:

165 = (-385 495 523 231 271 884)^3 + (383 344 975 542 639 445)^3 + (98 422 560 467 622 814)^3

795 = (-14 219 049 725 358 227)^3 + (2 337 348 783 323 923)^3 + (14 197 965 759 741 571)^3

906= (-74 924 259 395 610 397)^3 + (35 961 979 615 356 503)^3 + (72 054 821 089 679 353 378)^3

So the 8 remaining are:

114, 390, 579, 627, 633, 732, 921, and 975

Vince

Monday, April 4, 2022

________________________

I shared this email exchange with Vince because it is kind of typical of correspondence among mathematicians. Vince Chan is on faculty at the University of Calgary.

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