How do you solve this equation?

Algebra Level 3

Do there exist integers $x$ and $y$ satisfying the equation $x^3 + y^4 = 19^{19}$ ?

Cannot be determined Yes No

1 solution

Mark Hennings
Aug 1, 2018

We calculate that $x^3 \equiv 0,1,5,8,12 \pmod{13}$ , while $x^4 \equiv 0,1,3,9 \pmod{13}$ . Checking all the cases, we see that $x^3 + y^4 \equiv 0,1,2,3,4,5,6,8,9,10,11,12 \pmod{13}$ In fact, the only residue modulo $13$ that cannot be achieved by numbers of the form $x^3 + y^4$ is $7$ . Since $19^{19} \equiv 7 \pmod{13}$ , we see that the equation has no solutions.

What was the inspiration behind checking mod 13?

In general , I want to ask that in such type of diophantine equations , how do we come to know that what number modulo we have to check.

Vilakshan Gupta - 2 years, 10 months ago

The fact that $3$ and $4$ both divide $12=13-1$ means that there will be comparatively few third and fourth powers modulo $13$ , so that there was a chance that the sum of a cube and a fourth power might miss out a number modulo $13$ . It just worked.

This is an old problem, so I knew the solution. The above was the argument I first used to solve it (I first met it on a Number Theory course which was covering modulus arithmetic, so I knew that the trick was to find the right module to work with...)

Mark Hennings - 2 years, 10 months ago

How do you solve this equation?

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