How to Solve Diophantine Equations: Modulus

Diophantine equations are equations that are solved in integer numbers. We can solve some diophantine equations by considering the remainders of both sides of the equation by a certain modulus.

Problem (Australia, 1984)

Solve in integer numbers
$$ x^4+131=3y^4 $$

Solution

Answer: there are no such numbers.

Let us assume that such numbers exist and consider this equation modulo $5$. First let $n \in \mathbb{Z}$ and let us find all possible remainders of $n^4$ when divided by $5$:

 

if $n \equiv 0 \hspace{0.05in} (\text{mod } 5)$, then $n^4 \equiv 0^4 \equiv 0 \hspace{0.05in} (\text{mod } 5)$

if $n \equiv 1 \hspace{0.05in} (\text{mod } 5)$, then $n^4 \equiv 1^4 \equiv 1 \hspace{0.05in} (\text{mod } 5)$

if $n \equiv 2 \hspace{0.05in} (\text{mod } 5)$, then $n^4 \equiv 2^4 \equiv 1 \hspace{0.05in} (\text{mod } 5)$

if $n \equiv 3 \hspace{0.05in} (\text{mod } 5)$, then $n^4 \equiv 3^4 \equiv 1 \hspace{0.05in} (\text{mod } 5)$

if $n \equiv 4 \hspace{0.05in} (\text{mod } 5)$, then $n^4 \equiv 4^4 \equiv 1 \hspace{0.05in} (\text{mod } 5)$

 

 

This implies that $n^4$ is always congruent to $0$ or $1$ modulo $5$.
Notice that the left-hand side of the equation $ x^4+131$ is always congruent to $1$ or $2$ modulo $5$, while the right-hand side of the equation $ 3y^4$ is always congruent to $0$ or $3$ modulo $5$. Contradiction. Therefore such numbers do not exist.

---