Question

(0, 1, 2), (3, 0, 3) and (4, 2, 5) are only three solutions for the Diophantine equation 2 x + 3y = z 2 where x, y and z are non-negative integers.

prove it.​

Answer 1

Step-by-step explanation:

Proof. Let x, y and z be non-negative integers such that 2x + 3y = z

2

. We

will divide the number x into two cases.

Case 1. x is even. Let x = 2k where k is a non-negative integer. Then

4

k + 3y = z

2

. By Lemma 2.2, we have (k, y, z) ∈ {(0, 1, 2),(2, 2, 5)}. Thus,

(x, y, z) ∈ {(0, 1, 2),(4, 2, 5)}.

Case 2. x is odd. We will divide the number y into three subcases.

Subcase 2.1. y = 0. Then 2x + 1 = z

2

. By Lemma 2.3, we have x = 3 and

z = 3. Thus, (x, y, z) = (3, 0, 3).

Subcase 2.2. y is even and y > 0. Let y = 2l where l is a positive integer.

Then z

2−3

2l = 2x

. Then (z−3

l

)(z+3l

) = 2x

. Thus, z−3

l = 2u where u is a non-

negative integer. Then z + 3l = 2x−u

. Thus, 2(3l

) = 2x−u − 2

u = 2u

(2x−2u − 1).

Then u = 1. Then 3l = 2x−2 − 1. Then 2x−2 − 3

l = 1. Since l ≥ 1, we obtain

that 2x−2 = 3l + 1 ≥ 3

1 + 1 = 4. Then x ≥ 4. By Proposition 2.1, we have

l = 1. Then 2x−2 = 4. Then x = 4. This is a contradiction.

Subcase 2.3. y is odd. Note that z is also odd. Then z

2 ≡ 1 (mod 4). Let

y = 2l + 1 where l is a non-negative integer. Then z

2 = 2(4k

) + 3(9l

) ≡ 3 (mod

4). This is a contradiction.

Therefore, (0, 1, 2), (3, 0, 3) and (4, 2, 5) are only three solutions for the

Diophantine equation 2x + 3y = z

2 where x, y and z are non-negative integers.