$a^3 + b^4 = 19^{19}$

Question

We have to prove there doesn't exist such $a$ and $b$ that $$a^3 + b^4 = 19^{19} $$ I am totally stuck because I have tried numerous modulos and none of them works.

Thanks!

Answer 1

We want to limit the possible choices for $a^3$ and $b^4$ by using Fermat's Theorem. 13 is a good choice since then $x^{12} \equiv 1 \pmod {13}$ and 3 and 4 both divide 12.

First, work out $19^{19}$ mod 13. Hacking away at this, we find that it's equivalent to 7.

Now let's look at $a^3$. We know that $(a^3)^4 \equiv 1$, which means that $a^3$ must be some fourth root of 1 mod 13. If we play around a bit, we find that these are 1, -1, 5, and -5 (i.e. 1, 5, 8, and 12).

Similarly, for $b^4$, $(b^4)^3 \equiv 1$, so $b^4$ must be a cube root of 1 mod 13. After more playing around, we find these to be 1, 3, and 9.

Checking these carefully, we see that we're in luck - no combination gives us 7 mod 13.

Answer 2

According to PARI/GP, there is no solution modulo $13$.

? p=13;for(a=0,p-1,for(b=0,p-1,if(Mod(a^3+b^4-19^19,p)==0,print(a," ",b))))
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