How to prove for all xez x greater than or equal to 0 if x2 plus 4x plus 3 is prime implies x equal to 0?

Answer 1

I guess your question is:

∀ x ∈ ℤ with x ≥ 0, prove x² + 4x + 3 = p (where p is a prime) implies that x = 0.

Re-arrange and solve the quadratic:

x² + 4x + 3 = p

→ x² + 4x + (3 - p) = 0

→ x = (-4 ±√(4² - 4 × 1 × (3 - p)) / 2

→ x = -2 ±√(4 - 3 + p)

→ x = -2 ±√(p + 1)

As x ≥ 0, only the positive square root of (p + 1) is possible (otherwise x would be less than 0), thus:

x = √(p + 1) - 2

As x ∈ ℤ, p + 1 must be a perfect square (otherwise √(p + 1) is irrational and x ∉ ℤ).

Now, the difference of two squares is:

a² - b² = (a + b)(a - b)

If b = 1, this becomes:

a² - 1 = (a + 1)(a - 1)

→ a² = (a + 1)(a - 1) + 1

Thus all perfect squares are one more than the product of one less than their square root with one more than their square root.

All prime numbers have the factors 1 and themselves.

Thus, the only perfect square that is one more than a prime is when the square root is one more than 1, ie 1 + 1 = 2, giving:

2² = (2 + 1)(2 - 1) + 1 = 3 × 1 = 3 + 1 = a prime plus one.

(any higher perfect square is always one more than a compound number as:

a>2 gives: (a + 1) > 2, (a - 1) > 1 and (a + 1)(a - 1) is the product of two different numbers, neither of which is 1, thus the product is a compound number)

Therefore the only solution possible is when the perfect square is one more than the prime 3, ie:

x = √(p + 1) - 2 = √(3 + 1) - 2 = √4 - 2 = 2 - 2 = 0.

QED.

Answer 2

You factor the expression. Since, if x is a whole number, each of the factors is also a whole number, it follows that the expression is the product of two whole numbers, and therefore - if x is greater than 0 - not a Prime number.