Number theory class

Q. Find the value of

$$\frac{x^4 - 6x^3 - 2x^2 + 18x + 23}{x^2 - 8x + 15}$$

where

$$x = \sqrt{19 - 8\sqrt{3}}$$

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Solution:

Nice one! Let $x=\sqrt{19-8\sqrt3}$.
Recognize $19-8\sqrt3=(\sqrt{16}-\sqrt{3})^2$, so

$$x=4-\sqrt3.$$

Hence $x$ satisfies its quadratic:

$$(4-\sqrt3)^2=19-8\sqrt3 \;\Rightarrow\; x^2-8x+13=0 \quad\Rightarrow\quad x^2=8x-13.$$

Use this to reduce powers:

$$\begin{aligned} x^3&=x(8x-13)=8x^2-13x=8(8x-13)-13x=51x-104,\\ x^4&=x(51x-104)=51x^2-104x=51(8x-13)-104x=304x-663. \end{aligned}$$

Now the numerator:

$$\begin{aligned} x^4-6x^3-2x^2+18x+23 &=(304x-663)-6(51x-104)-2(8x-13)+18x+23\\ &=(304-306-16+18)x+(-663+624+26+23)\\ &=0\cdot x+10=10. \end{aligned}$$

And the denominator:

$$x^2-8x+15=(x^2-8x+13)+2=0+2=2.$$

Therefore,

$$\frac{x^4-6x^3-2x^2+18x+23}{x^2-8x+15}=\frac{10}{2}=\boxed{{\displaystyle 5}}\mathrm{.}$$

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Q. Find the unit digit of the expression

$$x={(\frac{-2a}{4+a}-\frac{\sqrt{\mathrm{\mid }a\mathrm{\mid }-3}+\sqrt{3-\mathrm{\mid }a\mathrm{\mid }}}{3-a})}^{1993}$$
Solution:
|a| - 3 >= 0 and 3 - |a| >= 0
=> a +-3
a = 3 will make denominator 0.
=> a = -3
x = (-6)^1993
Any power of an integer ending in 6 will be 6.
Answer = 6.

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Q. How many ordered pairs of +ve integers $(x, y)$ satisfy the equation

$$x\sqrt{y}+y\sqrt{x}+\sqrt{2006xy}-\sqrt{2006x}-\sqrt{2006y}-2006=0$$

Solution:

Let $a=\sqrt n$ and $b=\sqrt y$.
Then

$$n\sqrt y+y\sqrt n+\sqrt{2006ny}-\sqrt{2006n}-\sqrt{2006y}-2006 =ab(a+b)+\sqrt{2006}(ab-a-b)-2006.$$

Notice this factors nicely:

$$ab(a+b)+\sqrt{2006}(ab-a-b)-2006 =(ab-\sqrt{2006})(a+b+\sqrt{2006}).$$

Since $a,b,\sqrt{2006}>0$, the second factor is never $0$.
So we must have $ab=\sqrt{2006}$, i.e. $\sqrt{ny}=\sqrt{2006}$, hence $ny=2006$.

Thus the number of ordered positive integer pairs $(n,y)$ equals the number of divisors of $2006$.

$$2006=2\cdot17\cdot59 \quad\Rightarrow\quad d(2006)=(1+1)^3=8.$$

Answer: 8 ordered pairs (e.g., $(1,2006),(2,1003),(17,118),(34,59)$ and their swaps).

Q:

$$ \left(5 + 2\sqrt{6}\right)^{x^2 - 3} + \left(5 - 2\sqrt{6}\right)^{x^2 - 3} = 10  $$ Find x.

Solution:

Let $a=5+2\sqrt6$. Then $(5-2\sqrt6)=a^{-1}$ since $a(5-2\sqrt6)=1$.

So the equation becomes

$$a^{t}+a^{-t}=10,\quad \text{where } t=x^2-3.$$

Let $u=a^{t}$. Then $u+\frac1u=10 \Rightarrow u^2-10u+1=0$, giving

$$u=5\pm 2\sqrt6.$$

But $5+2\sqrt6=a$ and $5-2\sqrt6=a^{-1}$. Hence $a^{t}=a$ or $a^{t}=a^{-1}$, so $t=1$ or $t=-1$.

Thus $x^2-3=\pm1\Rightarrow x^2=4$ or $x^2=2$, and

$$\boxed{{\displaystyle x=\pm 2,\pm \sqrt{2}}}\mathrm{.}$$asdf