Extracurricular Elementary Mathematics Marathon (1 Viewer)

Paradoxica · Jan 21, 2016

Rules are as per the other marathons. Difficulty should be reasonable, and hints should be provided as deemed necessary. Use any "elementary" techniques, provided you state them, and if it's fairly advanced, outline a proof.

I'll start off simple.

$$Find all integer pairs $(x,y)$ that satisfy $x^2 + 6xy + 8y^2 +3x +6y = 2$

Other things:

For the purposes of this thread, the set of all natural numbers excludes zero.

Vectors and elementary functions of any kind are allowed, but little to no calculus. (this one due to leehuan)

Define terms that the average person following this thread probably wouldn't know.

State any theorems/techniques that may help in solving the problem.

InteGrand · Jan 21, 2016

Paradoxica said:
Rules are as per the other marathons. Difficulty should be reasonable, and hints should be provided as deemed necessary. Use any "elementary" techniques, provided you state them, and if it's fairly advanced, outline a proof.

I'll start off simple.

$$Find all integer pairs $(x,y)$ that satisfy $x^2 + 6xy + 8y^2 +3x +6y = 2$

$\noindent Note that the L.H.S. factorises into $(x+2y)(x+4y+3)$. This can be motivated by writing the L.H.S. as $x^2 + (6y+3)x+ \left(8y^2 + 6y\right)$, and factorising this quadratic in $x$. We want to write it as $(x+\alpha)(x+\beta)$, where $\alpha \beta = 8y^2 + 6y$ and $\alpha + \beta = 6y+3$. It is easy to see that we take $\alpha = 2y,\beta = 4y+3$.$

$\noindent So we have $(x+2y)(x+4y+3)=2$. Since $x,y\in \mathbb{Z}$, each bracketed term is an integer, so based on the factors of 2 being $2$ and $1$, we have the following four possibilities:$

$\noindent 1) $x+2y=2$ and $x+4y+3 = 1$$

$\noindent 2) $x+2y=-2$ and $x+4y+3 = -1$$

$\noindent 3) $x+2y=1$ and $x+4y+3 = 2$$

$\noindent 4) $x+2y=-1$ and $x+4y+3 = -2$.$

$\noindent In general, if(f) $x+2y=a$, then $x=a-2y$, so $x+4y+3 =b\Longleftrightarrow a-2y+4y+3 = b \Longleftrightarrow 2y = b-a -3 \Longleftrightarrow y=\frac{b-a-3}{2}$. This implies that $y$ will be integral iff $b-a$ is odd. This is the case in all the possibilities 1) to 4). So in 1), we have $y=\frac{1-2-3}{2}=\frac{-4}{2}=-2$, and $x=2-2\times (-2) = 6$.$

$\noindent In 2), we have $y=\frac{-1-(-2)-3}{2}=\frac{-2}{2}=-1$, so $x=-2-2\times (-1) = 0$.$

$\noindent In 3), we again have $y=\frac{2-1-3}{2}=\frac{-2}{2}=-1$, so $x=-2-2\times (-1) = 0$.$

$\noindent In 4), we again have $y=\frac{-2-(-1)-3}{2}=\frac{-4}{2}=-2$, so $x=-2-2\times (-2) = 6$.$

$\noindent So the solutions $(x,y)$ are $(6,-2)$ and $(0,-1)$.$

InteGrand · Jan 21, 2016

GoldyOrNugget · Jan 21, 2016

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Let $a$ and $b$ be real numbers with $a>1$ and $b\neq 0$. Suppose that $\frac{a}{b}=a^{3b}$ and $ab=a^{b}$. Find $b^{-a}$.$

16

InteGrand · Jan 21, 2016

GoldyOrNugget said:
16

Correct!

Paradoxica · Jan 21, 2016

InteGrand said:
$\textbf{NEW QUESTION}$

$\noindent Let $a$ and $b$ be real numbers with $a>1$ and $b\neq 0$. Suppose that $\frac{a}{b}=a^{3b}$ and $ab=a^{b}$. Find $b^{-a}$.$

The answer, has already been posted, but here, we will prove uniqueness.

$$\noindent$ \frac{a}{b} = (a^b)^3 = (ab)^3 = a^3b^3 \Leftrightarrow b^4 a^2 =1 \Leftrightarrow b^2 a = \pm 1 \\$But $a$ is a log-positive number, and $b$ is the result of a real-valued exponentiation, so $b$ is strictly non-negative, in the interval $(0,1)$. Hence, the only true relation is $b\sqrt{a} =1$. By substitution of $b$, we have $b^{-a} = a^{\frac{a}{2}}$. Lastly, observe that the function $y = x^{\frac{x}{2}}$ is injective on the interval $(1,\infty)$. This is sufficient to prove the uniqueness of $b^{-a}. \blacksquare$

Paradoxica · Jan 21, 2016

$$\noindent Let $p$ be an odd prime, $a$ be coprime to $p$, and $n$ be the smallest integer such that $\\a^n \equiv 1 $ mod $ p \\ $Show $n|(p-1)$

Hint: Use Fermat's Little Theorem.

InteGrand · Jan 21, 2016

Paradoxica said:
The answer, has already been posted, but here, we will prove uniqueness.

$$\noindent$ \frac{a}{b} = (a^b)^3 = (ab)^3 = a^3b^3 \Leftrightarrow b^4 a^2 =1 \Leftrightarrow b^2 a = \pm 1 \\$But $a$ is a log-positive number, and $b$ is the result of a real-valued exponentiation, so $b$ is strictly non-negative, in the interval $(0,1)$. Hence, the only true relation is $b\sqrt{a} =1$. By substitution of $b$, we have $b^{-a} = a^{\frac{a}{2}}$. Lastly, observe that the function $y = x^{\frac{x}{2}}$ is injective on the interval $(1,\infty)$. This is sufficient to prove the uniqueness of $b^{-a}. \blacksquare$

$\noindent It is probably easier to prove uniqueness (and get the actual answer) by just solving the given pair of simultaneous equations (fairly easy, just start by taking the log base $a$ of both sides of each equation $\ddot{\smile}$.).$

Paradoxica · Jan 21, 2016

InteGrand said:
$\noindent It is probably easier to prove uniqueness (and get the actual answer) by just solving the given pair of simultaneous equations (fairly easy, just start by taking the log base $a$ of both sides of each equation $\ddot{\smile}$.).$

That doesn't prove uniqueness of b, it just restricts the value range of b to the solution interval without proof...

Drsoccerball · Jan 21, 2016

Isn't this just the Advanced X2 Marathon ?

InteGrand · Jan 21, 2016

Paradoxica said:
That doesn't prove uniqueness of b, it just restricts the value range of b to the solution interval without proof...

$\noindent What do you mean? You find the actual value of $b$ by solving the equations using reversible steps (since there turns out to be exactly one solution, $b$ is unique). Taking logs of negative numbers though would be non-elementary so you'd maybe want to justify that $b$ is positive first. This is easily done by looking at the given equation $ab = a^b$ for example. The R.H.S. is positive, so $ab$ must be positive, so $b$ is positive as $a$ is given to be positive.$

Paradoxica · Jan 21, 2016

Drsoccerball said:
Isn't this just the Advanced X2 Marathon ?

No, it is unrestricted from the arbitrary bounds placed on it. I can't just post any Olympiad problem on there, but I can do so here.

leehuan · Jan 21, 2016

Drsoccerball said:
Isn't this just the Advanced X2 Marathon ?

If this were my thread I would allow hyperbolic functions and elementary vector notation now, however probably a bit of guidance as to how they work as we haven't attended a lecture yet.

Paradoxica · Jan 21, 2016

leehuan said:
If this were my thread I would allow hyperbolic functions and elementary vector notation now, however probably a bit of guidance as to how they work as we haven't attended a lecture yet.

Olympiad allows for that, but I've never seen it in good use. Except for vectors, those things are really useful for a good bunch of problems.

seanieg89 · Jan 21, 2016

Paradoxica said:
$$\noindent Let $p$ be an odd prime, $a$ be coprime to $p$, and $n$ be the smallest integer such that $\\a^n \equiv 1 $ mod $ p \\ $Show $n|(p-1)$

Hint: Use Fermat's Little Theorem.

p-1 = mn + r for non-negative integers m, r with r < n.

Then (a^r)(a^n)^m=a^(p-1)
=> a^r=1 (using FLT)

From the minimality of n, we must conclude r=0.

That is n|p-1.

seanieg89 · Jan 21, 2016

Solve the equation x^2+y^2+1=xyz over the positive integers.

Hint: First concentrate on determining what possibilities there are for z.

Paradoxica · Jan 22, 2016

seanieg89 said:
Solve the equation x^2+y^2=xyz over the positive integers.

Hint: First concentrate on determining what possibilities there are for z.

$$\noindent Divide down $xy$ without worry, as they are positive. Let $q = \frac{x}{y}$ be a rational number. We now obtain a quadratic in terms of $q,\; q^2 - qz +1 = 0$. By the quadratic formula, the discriminant must be a perfect square for $q$ to be rational. $z^2 -4 = k^2$ for some integer $k$. But the only valid integer solution for $z$ is 2, which means the only possible value of $z$ for our initial equation is 2. Finally, we have $x^2 + y^2 = 2xy$. But by the AM-GM inequality, this only happens if $x=y$. So $(r,r,2)$ is the only solution for any positive integer $r$.$

Paradoxica · Jan 22, 2016

GoldyOrNugget · Jan 22, 2016

Paradoxica said:
$$\noindent Show that there must exist a non-empty subset in a set of $n$ integers such that their sum is divisible by $n$.$

Trivial for n=1.

Suppose the property holds for 1..n-1.

For general n, if their sum is divisible by n, then their sum mod n is 0. Take all elements mod n. If all elements mod n are 0, pick any subset. Otherwise, pick out an element x mod N which is nonzero mod n. 0 <= x < n. Of the remaining elements, take any n-x > 0 elements. These n-x elements have a subset divisible by n-x, so if we then add x to the subset, we have a subset with sum (n-x) + x = 0 mod N, so the subset is divisible by n.

seanieg89 · Jan 22, 2016

Paradoxica said:
$$\noindent Divide down $xy$ without worry, as they are positive. Let $q = \frac{x}{y}$ be a rational number. We now obtain a quadratic in terms of $q,\; q^2 - qz +1 = 0$. By the quadratic formula, the discriminant must be a perfect square for $q$ to be rational. $z^2 -4 = k^2$ for some integer $k$. But the only valid integer solution for $z$ is 2, which means the only possible value of $z$ for our initial equation is 2. Finally, we have $x^2 + y^2 = 2xy$. But by the AM-GM inequality, this only happens if $x=y$. So $(r,r,2)$ is the only solution for any positive integer $r$.$

Nice

. I actually forgot a term on the LHS from the diophantine equation I was trying to remember though lol, try the edited problem too.

It is slightly harder, but not greatly so. The original hint still stands.

Repost for visibility:

Solve the equation:

$x^2+y^2+1=xyz$

over the positive integers.

Hint: Try to determine what z can be first.

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