Answer:
We can show that, regardless of the value of x and y, 3x² + 4y² is not a multiple of 5 (This is true if x and y are multiples of 5). We will see congruences module 5
First, lets see the congruence of x² module 5, which depends on the congruence of x module 5
- If the congruence of x module 5 is 0, then the congruence of x² module 5 is also 0
- If the congruence of x module 5 is 1, then the congruence of x² module 5 is 1² = 1
- If the congruence of x is 2, then the congruence of x² is 2² = 4
- If the congruence of x is 3, then the congruence of x² is 3² = 9-5 = 4
- if the congruence of x is 4, then the congruence of x² is 4² = 16-15 = 1
Therefore, the congruence of x² is always 0,1 or 4
The congruence of 3x² will always be 0, 3 or 2 (2 = 4*3-10)
The congruence of 4y² will always be 0,4 or 1 (1 = 4*4-15)
Note that the sum 3x² + 4y² is a multiple of 5 only when both 3x² and 4y², and thus, x and y are.
As a result, x = 5k, y = 5j, for certain values of k and j, and
3x²+4y² = 75k²+100j² = 5 * (15 k² + 20 j²) = 5z²
Therefore
15k² + 20j² = z²
So, z is a multiple of 5, 5 = 5l, and z² = (5l)² = 25 l². Our equation for k,j and l is reduced to
15 k² + 20j² = 25 l²
We can simplify it by dividing it by 5, obtaining
3k²+4j² = 5 l²
Which is equal to the original equation
This means that we could reduce the equation endlessly, which is a contradiction. Thus, 3x² + 4y² = 5z² has no trivial solution.
Let x = p/q and y = r/s be a rational solution of 3x² + 4y² = 5, then
3 p²/q² + 4 r²/s² = 5
multiplying by q²s² we have
3 p²s² + 4 r²q² = 5 q²s², or, equivalently,
3 (ps)² + 4(rq)² = 5(qs)²
Which only happens when ps=r1=qs = 0. Thus, q or s must be 0, which cant be true. As a result, the equation 3x² + 4y² = 5 has no rational solutions.
