# Extracurricular Elementary Mathematics Marathon (1 Viewer)

##### -insert title here-
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.

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

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.

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#### InteGrand

##### Well-Known Member
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.

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

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

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

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

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

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

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

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

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

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

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

#### InteGrand

##### Well-Known Member
$\bg_white \textbf{NEW QUESTION}$

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

#### GoldyOrNugget

##### Señor Member
$\bg_white \textbf{NEW QUESTION}$

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

Correct!

##### -insert title here-
$\bg_white \textbf{NEW QUESTION}$

$\bg_white \noindent Let a and b be real numbers with a>1 and b\neq 0. Suppose that \frac{a}{b}=a^{3b} and ab=a^{b}. Find b^{-a}.$
The answer, has already been posted, but here, we will prove uniqueness.

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

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

Hint: Use Fermat's Little Theorem.

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#### InteGrand

##### Well-Known Member
The answer, has already been posted, but here, we will prove uniqueness.

$\bg_white \noindent \frac{a}{b} = (a^b)^3 = (ab)^3 = a^3b^3 \Leftrightarrow b^4 a^2 =1 \Leftrightarrow b^2 a = \pm 1 \\But a is a log-positive number, and b is the result of a real-valued exponentiation, so b is strictly non-negative, in the interval (0,1). Hence, the only true relation is b\sqrt{a} =1. By substitution of b, we have b^{-a} = a^{\frac{a}{2}}. Lastly, observe that the function y = x^{\frac{x}{2}} is injective on the interval (1,\infty). This is sufficient to prove the uniqueness of b^{-a}. \blacksquare$
$\bg_white \noindent It is probably easier to prove uniqueness (and get the actual answer) by just solving the given pair of simultaneous equations (fairly easy, just start by taking the log base a of both sides of each equation \ddot{\smile}.).$

##### -insert title here-
$\bg_white \noindent It is probably easier to prove uniqueness (and get the actual answer) by just solving the given pair of simultaneous equations (fairly easy, just start by taking the log base a of both sides of each equation \ddot{\smile}.).$
That doesn't prove uniqueness of b, it just restricts the value range of b to the solution interval without proof...

#### Drsoccerball

##### Well-Known Member
Isn't this just the Advanced X2 Marathon ?

#### InteGrand

##### Well-Known Member
That doesn't prove uniqueness of b, it just restricts the value range of b to the solution interval without proof...
$\bg_white \noindent What do you mean? You find the actual value of b by solving the equations using reversible steps (since there turns out to be exactly one solution, b is unique). Taking logs of negative numbers though would be non-elementary so you'd maybe want to justify that b is positive first. This is easily done by looking at the given equation ab = a^b for example. The R.H.S. is positive, so ab must be positive, so b is positive as a is given to be positive.$

##### -insert title here-
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

##### Well-Known Member
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.

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##### -insert title here-
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

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

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

##### Well-Known Member
Solve the equation x^2+y^2+1=xyz over the positive integers.

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

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##### -insert title here-
Solve the equation x^2+y^2=xyz over the positive integers.

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

##### -insert title here-
$\bg_white \noindent Show that there must exist a non-empty subset in a set of n integers such that their sum is divisible by n.$

#### GoldyOrNugget

##### Señor Member
$\bg_white \noindent Show that there must exist a non-empty subset in a set of n integers such that their sum is divisible by n.$
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

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

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

over the positive integers.

Hint: Try to determine what z can be first.

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