# Homework Help: Diophantine Equations

1. Mar 11, 2006

### Oxymoron

Does anyone know of any resources on the web (or if you prefer, provide me directly with assistance) which will help me understand how to solve equations of the form

$$ax^3+by^3=c$$

I believe they are third-order Diophantine equations.

2. Mar 12, 2006

### Oxymoron

Perhaps I should be a little more specific.

How would you go about proving that a third order Diophantine equation has no solutions?

3. Mar 14, 2006

### Oxymoron

I decided to have a go and here is how I went. Let me know if I made any mistakes

Consider the Diophantine equation

$$x^3+117y^3=5$$

Choose mod 5.

The equation tells us that $5|x^3 \Rightarrow 5^3|x^3 \Rightarrow x = 5X$. Therefore

$$125\cdot X^3 +117y^3 = 5$$

By the same procedure as above, this equation tells us that $5|y^3 \Rightarrow y = 5Y$. Therefore

$$125\cdot X^3 + 117\cdot 5Y^3 = 5$$

Divide through by 5 and we have

$$25\cdot X^3 + 117\cdot Y^3 = 1$$

This equation now tells me that

$$25X^3 \equiv 1(\mod 5)$$

But $25\equiv 0 (\mod 5)$. Hence there is no such $0,1,2,3,4$ such that $X^3\equiv 1(\mod 5)$.

However, if we had checked Y first we would have found

$$117Y^3 \equiv 1(\mod 5)$$

which implies that

$$Y^3 \equiv 2(\mod 5)$$

since $117\equiv 2 (\mod 5)$. And since if we let $Y = 3$ then

$$Y^3 = 3^3 = 27 \equiv 2(\mod 5)$$

then this tells us that there is a solution. But since it failed for X there is no solution.