Question

The number of integral values of x for which $2x^2 + x - 6$ is a positive integral power of a positive prime is

Answer 1

3

Answer 2

Let the given expression be $E = 2x^2 + x - 6$. We are looking for integral values of $x$ such that $E$ is a positive integral power of a positive prime.

First, factorize the quadratic expression: $2x^2 + x - 6 = 2x^2 + 4x - 3x - 6 = 2x(x+2) - 3(x+2) = (2x-3)(x+2)$.

So, we need $(2x-3)(x+2) = p^k$, where $p$ is a positive prime and $k$ is a positive integer. Since $x$ is an integer, both $(2x-3)$ and $(x+2)$ are integers. For their product to be a positive power of a prime, both factors must be powers of the same prime $p$, and they must have the same sign (both positive or both negative).

Let $A = x+2$ and $B = 2x-3$. So $AB = p^k$. We can express $B$ in terms of $A$: $2A = 2(x+2) = 2x+4$. Then $2A - B = (2x+4) - (2x-3) = 7$. So, we have the equation $2A - B = 7$, where $A=p^b$ and $B=p^a$ for some non-negative integers $a, b$ (since $A$ and $B$ are factors of $p^k$). Thus, $2p^b - p^a = 7$.

Case 1: Both factors $A$ and $B$ are positive. This implies $x+2 > 0 \implies x > -2$ and $2x-3 > 0 \implies x > 3/2$. So, for integer $x$, we must have $x \ge 2$. In this case, $p^a > 0$ and $p^b > 0$.

We need to solve $2p^b - p^a = 7$ for prime $p$ and non-negative integers $a, b$.

Subcase 1.1: $p=2$. The equation becomes $2 \cdot 2^b - 2^a = 7$, which simplifies to $2^{b+1} - 2^a = 7$. If $a=0$, then $2^{b+1} - 1 = 7 \implies 2^{b+1} = 8 = 2^3$. This gives $b+1=3 \implies b=2$. So, $p=2, a=0, b=2$. $A = x+2 = p^b = 2^2 = 4 \implies x=2$. $B = 2x-3 = p^a = 2^0 = 1$. Check: $(2x-3)(x+2) = (1)(4) = 4 = 2^2$. This is a positive integral power of a positive prime (2). So $x=2$ is a solution.

If $a \ge 1$: $2^a(2^{b+1-a} - 1) = 7$. Since $7$ is a prime number, $2^a$ must be a factor of 7. The only positive integer factor of 7 that is a power of 2 is $2^0=1$. So $a$ must be 0, which we already covered. Thus, for $p=2$, the only solution is $x=2$.

Subcase 1.2: $p=3$. The equation becomes $2 \cdot 3^b - 3^a = 7$. If $a=0$, $2 \cdot 3^b - 1 = 7 \implies 2 \cdot 3^b = 8 \implies 3^b = 4$. No integer $b$ satisfies this. If $b=0$, $2 \cdot 1 - 3^a = 7 \implies 3^a = -5$. No solution. If $a=1$, $2 \cdot 3^b - 3 = 7 \implies 2 \cdot 3^b = 10 \implies 3^b = 5$. No solution. If $b=1$, $2 \cdot 3 - 3^a = 7 \implies 6 - 3^a = 7 \implies 3^a = -1$. No solution. If $a \ge 1$, $3^a(2 \cdot 3^{b-a} - 1) = 7$. $3^a$ must be 1, so $a=0$, which we already covered. No solutions for $p=3$.

Subcase 1.3: $p=5$. The equation becomes $2 \cdot 5^b - 5^a = 7$. If $a=0$, $2 \cdot 5^b - 1 = 7 \implies 2 \cdot 5^b = 8 \implies 5^b = 4$. No solution. If $b=0$, $2 \cdot 1 - 5^a = 7 \implies 5^a = -5$. No solution. No solutions for $p=5$.

Subcase 1.4: $p=7$. The equation becomes $2 \cdot 7^b - 7^a = 7$. Divide by 7: $2 \cdot 7^{b-1} - 7^{a-1} = 1$. For this equation to hold, either $a-1=0$ or $b-1=0$, or both are greater than 0. If $a-1=0 \implies a=1$: $2 \cdot 7^{b-1} - 7^0 = 1 \implies 2 \cdot 7^{b-1} - 1 = 1 \implies 2 \cdot 7^{b-1} = 2 \implies 7^{b-1} = 1 = 7^0$. This gives $b-1=0 \implies b=1$. So, $p=7, a=1, b=1$. $A = x+2 = p^b = 7^1 = 7 \implies x=5$. $B = 2x-3 = p^a = 7^1 = 7$. Check: $(2x-3)(x+2) = (7)(7) = 49 = 7^2$. This is a positive integral power of a positive prime (7). So $x=5$ is a solution.

If $b-1=0 \implies b=1$: $2 \cdot 7^0 - 7^{a-1} = 1 \implies 2 - 7^{a-1} = 1 \implies 7^{a-1} = 1 = 7^0$. This gives $a-1=0 \implies a=1$. This leads to the same solution $x=5$.

If $a-1 > 0$ and $b-1 > 0$: $7^{a-1}(2 \cdot 7^{b-a} - 1) = 1$ (assuming $a \le b$) or $7^{b-1}(2 - 7^{a-b}) = 1$ (assuming $b < a$). For the product to be 1, both factors must be 1. $7^{a-1}=1 \implies a-1=0 \implies a=1$. $2 \cdot 7^{b-a} - 1 = 1 \implies 2 \cdot 7^{b-a} = 2 \implies 7^{b-a} = 1 \implies b-a=0 \implies b=a$. This again leads to $a=b=1$. Thus, for $p=7$, the only solution is $x=5$.

Subcase 1.5: For $p > 7$. The equation is $2p^b - p^a = 7$. If $a=0$, $2p^b - 1 = 7 \implies 2p^b = 8 \implies p^b = 4$. Since $p$ is a prime, $p$ must be 2. But we assumed $p>7$. No solutions. If $b=0$, $2 - p^a = 7 \implies p^a = -5$. No solution. If $a \ge 1$ and $b \ge 1$: If $a=b$, $2p^a - p^a = 7 \implies p^a = 7$. This means $p=7$ and $a=1$, which we already found ($x=5$). If $a < b$, $p^a(2p^{b-a} - 1) = 7$. Since $p$ is prime, $p^a$ must be 1 or 7. If $p^a=1$, then $a=0$, which we already covered. If $p^a=7$, then $p=7$ and $a=1$, which we already found. If $a > b$, $p^b(2 - p^{a-b}) = 7$. Since $p$ is prime, $p^b$ must be 1 or 7. If $p^b=1$, then $b=0$, which we already covered. If $p^b=7$, then $p=7$ and $b=1$, which we already found. So, there are no new solutions for $p>7$ when $A, B > 0$.

Case 2: Both factors $A$ and $B$ are negative. This implies $x+2 < 0 \implies x < -2$ and $2x-3 < 0 \implies x < 3/2$. So, for integer $x$, we must have $x \le -3$. Let $A = -p^b$ and $B = -p^a$ for some non-negative integers $a, b$. The equation $2A - B = 7$ becomes $2(-p^b) - (-p^a) = 7 \implies -2p^b + p^a = 7 \implies p^a - 2p^b = 7$.

Subcase 2.1: $p=2$. $2^a - 2 \cdot 2^b = 7 \implies 2^a - 2^{b+1} = 7$. If $b=0$, $2^a - 2 = 7 \implies 2^a = 9$. No integer $a$. If $b=1$, $2^a - 4 = 7 \implies 2^a = 11$. No integer $a$. If $b=2$, $2^a - 8 = 7 \implies 2^a = 15$. No integer $a$. If $b=3$, $2^a - 16 = 7 \implies 2^a = 23$. No integer $a$. In general, $2^a - 2^{b+1} = 7$. Since $2^a$ and $2^{b+1}$ are powers of 2, their difference can only be 7 if one of them is small. If $a=0$, $1 - 2^{b+1} = 7 \implies 2^{b+1} = -6$. No solution. If $a=1$, $2 - 2^{b+1} = 7 \implies 2^{b+1} = -5$. No solution. If $a=2$, $4 - 2^{b+1} = 7 \implies 2^{b+1} = -3$. No solution. If $a=3$, $8 - 2^{b+1} = 7 \implies 2^{b+1} = 1 \implies b+1=0 \implies b=-1$. Not a non-negative integer. If $a \ge 1$: $2^{b+1}(2^{a-(b+1)} - 1) = 7$ (assuming $a > b+1$) or $2^a(1 - 2^{b+1-a}) = 7$ (assuming $a < b+1$). If $a > b+1$: $2^{b+1}$ must be 1, so $b+1=0 \implies b=-1$. Not allowed. If $a < b+1$: $2^a$ must be 1, so $a=0$. Not allowed. No solutions for $p=2$ in this case.

Subcase 2.2: $p=3$. $3^a - 2 \cdot 3^b = 7$. If $b=0$, $3^a - 2 = 7 \implies 3^a = 9 = 3^2$. This gives $a=2$. So, $p=3, a=2, b=0$. $A = x+2 = -p^b = -3^0 = -1 \implies x=-3$. $B = 2x-3 = -p^a = -3^2 = -9$. Check: $(2x-3)(x+2) = (-9)(-1) = 9 = 3^2$. This is a positive integral power of a positive prime (3). So $x=-3$ is a solution.

If $b \ge 1$: $3^b(3^{a-b} - 2) = 7$. Since $7$ is prime, $3^b$ must be 1. So $b=0$, which we already covered. No new solutions for $p=3$.

Subcase 2.3: For $p > 3$. The equation is $p^a - 2p^b = 7$. If $b=0$, $p^a - 2 = 7 \implies p^a = 9$. This implies $p=3$ and $a=2$, which we already found ($x=-3$). But we assumed $p>3$. No solutions. If $a=0$, $1 - 2p^b = 7 \implies 2p^b = -6 \implies p^b = -3$. No solution. If $a=b$, $p^a - 2p^a = 7 \implies -p^a = 7 \implies p^a = -7$. No solution. If $a > b$: $p^b(p^{a-b} - 2) = 7$. $p^b$ must be 1 or 7. If $p^b=1$, then $b=0$, which we already covered. If $p^b=7$, then $p=7$ and $b=1$. Then $p^{a-b} - 2 = 1 \implies 7^{a-1} = 3$. No integer $a$. If $a < b$: $p^a(1 - 2p^{b-a}) = 7$. $p^a$ must be 1 or 7. If $p^a=1$, then $a=0$, which we already covered. If $p^a=7$, then $p=7$ and $a=1$. Then $1 - 2p^{b-a} = 1 \implies 2p^{b-a} = 0$. No solution. No new solutions for $p>3$ in this case.

Combining all solutions for $x$: From Case 1 (positive factors): $x=2$ and $x=5$. From Case 2 (negative factors): $x=-3$.

The integral values of $x$ are $-3, 2, 5$. The number of integral values of $x$ is 3.

The final answer is $\boxed{3}$.