6. Prove that the diophantine equation 3x 2 + 4y 2 = 5z 2 has no non-trivial (i.e. (x, y, z) (0,0,0)) solutions. [Give a proof by contradict

Question

6. Prove that the diophantine equation 3x 2 + 4y 2 = 5z 2 has no non-trivial (i.e. (x, y, z) (0,0,0)) solutions. [Give a proof by contradiction. Obtain a contradiction by proving that if there is a non-trivial solution then there is a solution (x1 , y1 , z1 ) with x1 0 mod 5 or y1 0 mod 5.] Deduce that the equation 3x 2 + 4y 2 = 5 has no rational solutions.

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2 hours 2021-09-13T01:41:00+00:00 1 Answer 0

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.