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Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.

In addition, $b$ must be a natural number, which means that:

$1 + 2\cdot a\cdot c \ge 4$

$2\cdot a \cdot c \ge 3$

$a\cdot c \ge \frac{3}{2}$

But the lowest possible product made by two prime numbers is $2^{2} = 4$ . Hence, $a\cdot c \ge 4$ .

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Example: $a = 2$ , $c = 2$

$b = \sqrt{1 + 2\cdot (2)\cdot (2)}$

$b = 3$

4 0

3 years ago

Please help? Graph <img src="https://tex.z-dn.net/?f=-3x%5E2%2B12y%5E2%3D84" id="TexFormula1" title="-3x^2+12y^2=84" alt="-3x^2+

mariarad [96]

$-3x^2+12y^2=84$
$12y^2=3x^2+84$
$y^2=\dfrac{x^2}4+7$
$y=\pm\sqrt{\dfrac{x^2}4+7}$

For either square root to exist, you require that $\dfrac{x^2}4+7\ge0$ . This is true for all $x$ , since $\dfrac{x^2}4$ is always non-negative. This means the domain of $y$ as a function of $x$ is all real numbers, or $x\in\mathbb R$ or $(-\infty,infty)$ .

Now, because $\dfrac{x^2}4+7$ is non-negative, and the smallest value it can take on is 7, it follows that the minimum value for the positive square root must be $\sqrt7$ , while the maximum value of the negative root must be $-\sqrt7$ . This means the range is $y\in\mathbb R\setminus(-\sqrt7,\sqrt7)$ , or $|y|\ge\sqrt7$ , or $(-\infty,-\sqrt7]\cup[\sqrt7,\infty)$ .