Product of two even integers is even. For all integers n, n^2 + 5n + 6 is prime. √α + β = √α + √β. For any integer p, if p is prime then p^2 - 1 is even. For any odd integer n, 7n^... Product of two even integers is even. For all integers n, n^2 + 5n + 6 is prime. √α + β = √α + √β. For any integer p, if p is prime then p^2 - 1 is even. For any odd integer n, 7n^2 + 3 is even. If 3n + 7 is odd, then n is odd. For any integer n > 3, 2^n - 1 is prime. Read more

Steps to Solve

1. Evaluate product of two even integers The product of two even integers $2m$ and $2n$ can be expressed as: $$ (2m)(2n) = 4mn $$ Since $4mn$ is divisible by 2, it is even.

2. Test for the polynomial being prime The expression $n^2 + 5n + 6$ factors to $(n + 2)(n + 3)$. This is not prime for most integer values. For example, for $n=0$, it equals $6$, which is not prime.

3. Check the square root equality The equality $\sqrt{\alpha} + \beta = \sqrt{\alpha} + \sqrt{\beta}$ holds if $\sqrt{\beta} = \beta$, which is true only if $\beta = 0$ or $\beta = 1$. Hence, this is generally false.

4. Evaluate $p^2 - 1$ for primes If $p$ is prime and greater than 2, then $p$ is odd, and $p^2$ is odd. Thus, $p^2 - 1$ is even.

5. Check if $7n^2 + 3$ is even for odd $n$ For odd integers $n = 2k + 1$, substituting gives: $$ 7(2k + 1)^2 + 3 = 7(4k^2 + 4k + 1) + 3 = 28k^2 + 28k + 10 $$ This is always even.

6. Determine if $3n + 7$ being odd implies $n$ is odd If $3n + 7$ is odd, $3n$ must be even, thus $n$ is even since the product of an odd number (3) and an odd integer ($n$) results in an odd product.

7. Test for $2^n - 1$ being prime for $n > 3$ Values like $n = 4$ give $2^4 - 1 = 15$, which is not prime. Thus, the statement is false for integers greater than 3.