# Fermat's Last Theorem - or is it?

Is there a solution in distinct positive integers $a,b,c$ to the equation $$a^3+b^3=c^4$$? If so, construct one; if not, prove that it can't possibly exist.

Don't be too put off by the appearance of this puzzle: it's nowhere near as hard as its famous relative Fermat's Last Theorem. There should be an AHA moment when you realise what you need, and the solution is very surprising if you haven't seen the like before.
This puzzle was discussed by Adam McBride in “Mathematics: The Greatest Subject in the World,” The Mathematical Gazette, vol. 89, no. 516 [November 2005].

• I have discovered a truly marvellous proof of this, which this comment is too narrow to contain.
– null
Commented Aug 17, 2015 at 12:41
• This is answered already, but also worth noting is <a href="bealconjecture.com/">Beal's conjecture</a> ($\$1,000,000$prize for solution!): if$A^x + B^y = C^z$has a solution in integers, then$A$,$B$,$C$must have some common factor. This is the case in all the examples given above. – user15764 Commented Aug 17, 2015 at 13:10 • Curse my math training. Every time I see an odd exponent my mind immediately jumps to negative numbers to test them. Commented Aug 18, 2015 at 20:03 • @null is that a reference to how Pierre de Fermat said how "he had the proof of his last theorem, but it was too long to fit"? Hahah :P Commented Mar 22, 2018 at 4:14 ## 2 Answers Trivial Solution$2^3+2^3=2^4$More Generally- Let$k=m^3+n^3\ \ where\ \ n>m>1$Then,$(mk)^3+(nk)^3=(m^3+n^3)k^3= k^4$For example, if$m=2, n=4$, we have$k=72$and$ 144^3 + 288^3 =72^4 $• My apologies - I forgot to exclude the trivial case. Check the edited OP. Commented Aug 17, 2015 at 8:11 • you basically gave the answer but notice that$a$and$c$are not distinct. you just need to pick both$m$and$n$larger than$1$. The smallest you can get then is$m=2$and$n=3$giving exactly Curtis' answer – Ivo Commented Aug 17, 2015 at 10:19 • @IvoBeckers Thanks!! I have a bad habit of overlooking small details. Commented Aug 17, 2015 at 10:40 • A couple of comments: (1) There's no reason to exclude$m=n$or to exclude 1. This should be "Let$k=m^3+n^3$where$n \ge m \ge 1$". (2) There is an even more general rule. For example,$417^3+1807^3=278^4$, but this doesn't fit the pattern described in this answer. Commented Aug 18, 2015 at 8:47 • @DavidHammen (1) I did include them first but the question asks for distinct solutions, the constraints were edited in for that purpose. See Ivo Beckers' comment above (2) I do have a little more generalized solution (though it is not a complete set of solutions) based on gcds, but the question only asks for a construction and what I showed is enough for that purpose. I did consider showing the generalized version, but IMO it really isn't much upgrade on the current one and I decided to keep it simple. Commented Aug 18, 2015 at 9:05 Same method as Anachor, pretty much:$70^3 + 105^3 = (2*35)^3 + (3*35)^3 = (2^3 + 3^3)*35^3 = 35^4\$

• I think you were the first to get a valid example, even though Anachor got the general method first ... so who should I give the tick to? Commented Aug 17, 2015 at 21:20
• Probably Anachor; his generates mine, given the right input. Commented Aug 19, 2015 at 14:14