Some properties of ferrous nitride

by Ankoganit, Jun 26, 2017, 3:03 PM

Ferrous nitride is a crystalline, metallic solid, usually found in powdered form, which is insoluble in water. It has the chemical formula $\text{Fe}_3\text{N}_2$.

Heh, just kidding; I won't tax you with these stuff. Chemistry was never really my forte anyway. What I am going to do is, though, to dwell on that neat formula $\text{Fe}_3\text{N}_2$. Doesn't it look cute? (Hint: it doesn't.) In honour of that elegant formula that allegedly holds the key to understanding the mysteries of our universe $^{\color{blue}[\textit{\textsf{citation needed}}]}$, I'll write the solutions (plus some motivation) of $3$ FEs and $2$ Number theory problem. Yeah, that's totally how I interpret chemical formulas. Guess who botched the chemistry exam last year?

Chemical Analysis #1: Surjection curing dejection
Quote:
(Iran TST 2011 P5) Find all surjective functions $f: \mathbb R \to \mathbb R$ such that for every $x,y\in \mathbb R,$ we have
\[f(x+f(x)+2f(y))=f(2x)+f(2y).\]

Darn, this is bad. Everything is inside those $f$'s; I doubt we can pull out anything out of them. And there's also that surjectivity thing; what can it do, anyway? Well, surjectivity allows us write $x=f(y)$ for any $x$ we wish...hang on, did we just pull out $x$ from $f$? Excellent! I bet this trick is going to be useful.

OK, let's start with plugging random values, 'cause that's pretty much all we can do at this stage. To make life easier, let $P(x,y)$ mean $P$lug in $x$ and $y$ into the given equation. So, we can plug in $0$ anywhere; we can even cook up a mystery constant $c$ so that $f(c)=0$ via surjectivity, that'd let us cancel a bunch of stuff. Elementary combinatorics (yay!) says there are basically $3\times 3$ plug-ins that might be useful, but we're not gonna try all of them, are we? Let's trying spamming everything with $c$'s, that'd get us a whole lot of zeros:$$P(c,c)\implies f(c)=2f(2c)\implies f(2c)=0.$$Cool, so even $2c$ leads to $0$. Can we use that fact anywhere? Of course, we can spam one of the variable slots with $c$, like so $$P(x,c)\implies f(f(x)+x)=f(2x)\quad (\star); \;\;P(c,x)\implies f(c+2f(x))=f(2x)\quad (\spadesuit).$$Now we're talking. If we can somehow manage to get injectivity, we can argue $f(f(x)+x)=f(c+2f(x))\implies f(x)=x-c$, and we'll be left with checking the linear functions. So we need to get hold of injectivity.

Well, let's begin just as all canonical injectivity proofs begin. Suppose $f(a)=f(b)$; we hope to prove that $a=b$ somehow. However we try to use the given equations, we'd run into things like $f(2a)$; so let's settle that. From $(\spadesuit)$, we have $f(2a)=f(c+2f(a))=f(c+2f(b))=f(2b)$. That's neat.

Now that we've got more letters to shuffle around, let's do that. $P(a,x)$ and $P(b,x)$ give $$f(2f(x)+a+f(a))=f(2x)+f(2a)=f(2x)+f(2b)=f(2f(x)+b+f(b))=f(2f(x)+b+f(a)).$$Now because of the surjectivity of $f(x)$, the expression $2f(x)+f(a)$ can take any real value, so $f(x+a)=f(x+b)\implies f(x+a-b)=f(x)$. Calling $k=a-b$, we have that $k$ is a period of $f$, and we wish to show $k=0$. But that'd require pulling $k$ out of the $f$'s somehow...of course, that's what the trick in the first paragraph was for! Let $\kappa$ be such that $f(\kappa)=k$, then we can use the periodicity of $f$, with $(\star)$ and $(\spadesuit)$ like so: $$k=f(\kappa)=f(\kappa+k)=f(\kappa+f(\kappa))=f(2\kappa)=f(c+2f(\kappa))=f(c+2k)=f(c)=0.$$As we've seen before, this leads us to $f(x)=x-c$, and now it's trivial to check that only $c=0$ works. And that seals the deal. $\blacksquare$

Chemical Analysis #2: Cubes versus newbs
Quote:
(ELMO 2017 P6) Find all functions $f:\mathbb{R}\to \mathbb{R}$ such that for all real numbers $a,b,$ and $c$:
  1. If $a+b+c\ge 0$ then $f(a^3)+f(b^3)+f(c^3)\ge 3f(abc).$
  2. If $a+b+c\le 0$ then $f(a^3)+f(b^3)+f(c^3)\le 3f(abc).$

Proposed by Ashwin Sah

So I finally managed to nail a P6. GG.

The only things that look somewhat scary are the fact that there are inequalities, not equalities, and that things are raised to third powers. We can squeeze out an equality out of the inequalities, but the cubes are going to be painful.

The first few steps are fairly routine: plug in zeros, $-x$, and all that stuff. Here's how it goes:
If $f$ is a function satisfying the conditions, then so is $f+k$ for any constant $k$, so we can assume WLOG $f(0)=0$. Note that the given statements imply:

$``$For reals $a,b,c$ with $a+b+c=0$, we have $f(a^3)+f(b^3)+f(c^3)=3f(abc)."$


Call this statement $P(a,b,c)$. Now $P(a^{\frac13},-a^{\frac13},0)$ gives $f(a)+f(-a)=0$, so $f$ is odd. Now for $a>0$, we have $a^{\frac13}+0+0\ge 0$, so condition $(i)$ gives $f(a)\ge 0$. Because of oddness, we have $f(a)\le 0$ for $a<0$. So $f$ takes nonnegative values at non-negative inputs and non-positive values for non-positive inputs.

Now we run into an impasse. First of all we need equalities, not inequalities; so the quoted statement would probably be the only useful thing from this point. Also, that $a+b+c=0$ is an annoying constraint; let's drop $c$ altogether:
Consider that statement $P(a,b,-(a+b))$. Because $f$ is odd, this translates into$$f(a^3)+f(b^3)+3f(ab(a+b))=f((a+b)^3).$$
Good work, but not great. There's nothing outside $f$'s, we lost a variable, and we're nowhere near completion. We need one more variable, at least. Where do we get that from? We introduce a new one, of course! This brings us to the next part of the solution:
Using the above equation multiple times, we obtain
\begin{align*} f((a+b+c)^3) &=f((a+b)^3)+f(c^3)+3f((a+b)c(a+b+c))\\ &= f(a^3)+f(b^3)+3f(ab(a+b))+f(c^3)\\ &+3f((a+b)c(a+b+c))\end{align*}Similarly, switching the roles of $b$ and $c$ we get, $$f((a+b+c)^3)=f(a^3)+f(b^3)+f(c^3)+3f(ac(a+c))+3f((b(a+c)(a+b+c)).$$Comparing these two expressions for $f((a+b+c)^3)$, we obtain $$f(c(a+b)(a+b+c))+f(ab(a+b))=f(b(a+c)(a+b+c))+f(ac(a+c)).\qquad (\star)$$
Ugh, that looks horrible. But that does hint at something; $f$ mostly respects addition, because if we ignored the $f$'s and added the terms in the above equation, we'd get an identity. The respect is shown in a rather ugly way, though. We need to ensure that those weird expressions can be replaced by generic $x,y$. So we need to find $a,b,c$ so that $c(a+b)(a+b+c)=x,ab(a+b)=y,b(a+c)(a+b+c)=z$. Oops, that's hard, mostly because it's a system of $3$ cubic equations. Let's loosen up the conditions a bit; we don't need $z$, just gotta make sure that $ab(a+b)=y=b(a+c)(a+b+c)$. Playing around with the equations a bit, noticing the homogeneity and some luck gets us to the following chain of reasoning:
Now choose arbitrary positive reals $x\le y$ and consider the following system of equations
\begin{align} a^2c-abc&=a^2b+ab^2\\ a^2b+ab^2&=x\\ a^2c+ac^2&=y \end{align}We claim that this has a solution in reals $a,b,c$. Indeed, set $b=qa,c=ra$, and $(1)$ becomes $$r-qr=q+q^2\implies r=\frac{q^2+q}{1-q}.$$And $(2)\div (3)$ becomes $$\frac{x}{y}=\frac{q(q+1)}{r(r+1)}=\frac{q(q+1)}{\frac{q(q+1)}{1-q}\cdot\frac{q^2+1}{1-q}}=\frac{(q-1)^2}{1+q^2}.$$The function $h(q)=\tfrac{(q-1)^2}{1+q^2}$ is continuous on $[0,1]$ and $h(0)=1,h(1)=0$. Since $\tfrac xy\in(0,1]$, there is a $q\in[0,1)$ so that $h(q)=\tfrac xy$. Choose this $q$, and the corresponding $r=\tfrac{q(q+1)}{1-q}$. Then scale both $q$ and $r$ by $a$ to get $b,c$ so that $(2)$ holds; then $(3)$ and $(1)$ would hold automatically. Thus $a,b,c$ exist; clearly $a\ne 0$.

Now that we've chosen suitable $a,b,c$, note that $$a^2c-abc=a^2b+ab^2\implies b(a+b+c)=ac\implies b(a+c)(a+b+c)=ac(a+c)=y,$$and $$ab(a+b)=x,$$So $$c(a+b)(a+b+c)=(b(a+c)(a+b+c)+ac(a+c))-ab(a+b)=2y-x.$$Using these in $(\star)$ gives $$f(2y-x)+f(x)=2f(y)\; \forall 0<x\le y.$$Setting $z=2y-x\iff y=\frac{x+z}{2}$, this becomes $$f(x)+f(z)=2f\left(\frac{x+z}{2}\right)\;\forall 0<x\le z.$$So $f$ satisfies Jensen's functional equation over positives, and it's bounded on $[0,\infty)$, so $f(x)=cx+d$ for some $c\ge 0$, which works.
Whoopee. $\blacksquare$

Chemical Analysis #3: Continuous f(rustration)
Quote:
(Tuymaada 2003, P4) Find all continuous functions $f(x)$ defined for all $x>0$ such that for every $x$, $y > 0$
\[ f\left(x+{1\over x}\right)+f\left(y+{1\over y}\right)= f\left(x+{1\over y}\right)+f\left(y+{1\over x}\right) . \]

Proposed by F. Petrov

I tried this months ago, without much success. I tried it again today morning, and having solved it, it doesn't seem even that hard. Possibly, I've actually messed up and managed to overlook some glaring flaw; so read at your own discretion, and let me know if it's glitched.

There's a lot of reciprocal business going on, so let's start with something like that. Replace $x$ by $\frac 1x$ in the given equation, and we have$$f\left(x+\frac1x\right)+f\left(y+\frac1y\right)=f\left(x+y\right)+f\left(\frac1x+\frac1y\right).$$
Notice something: $f$ respects addition in the same as the previous problem did; ignore the $f$'s and you end up with an identity. Now the key piece of intuition is: the pair $(x+y,\tfrac1x+\tfrac 1y)$ has the same sum as the $(x+\tfrac1x,y+\tfrac1y)$, but the numbers in the latter pair are closer together (check it!). This suggests some approach like this: take $a,b$, and express them as $x+y$ and $\tfrac 1x+\tfrac1y$, use the above equation and find a new pair of numbers $x+\tfrac1x$ and $y+\tfrac1y$ and make our new $a,b$, which are closer together. If we repeat this ad infinitum, the numbers in the pair will become equal in the limit (here's where we use continuity), and we can infer something nice. This motivates the following:

Take arbitrary positive reals $a,b$ so that $ab\ge 4$, and define the sequences $\left\{a_i\right\}_{i=0}^\infty,\left\{b_i\right\}_{i=0}^\infty$ as follows:
$$a_0=a,b_0=b,$$and for $n\ge 0$, take positive reals $x,y$ so that $x+y=a_n,\tfrac1x+\tfrac1y=b_n$ (these exist because $ab\ge 4$), and define $$a_{n+1}=x+\frac1x,b_{n+1}=y+\frac1y.$$
Because of the equation above, $$f(a_0)+f(b_0)=f(a_1)+f(b_1)=\cdots=f(a_n)+f(b_n)=\cdots \quad (\star).$$Also, $a_n+b_n$ is a constant throughout. Now take any $n$, and consider the numbers $a_n,b_n,a_{n+1},b_{n+1}$. Let $x,y$ be the corresponding values that satisfy $x+y=a_n,\tfrac1x+\tfrac1y=b_n.$ This implies $xy=\tfrac{a_n}{b_n}.$ Then we have $$\frac{|a_{n+1}-b_{n+1}|}{|a_n-b_n|}=\left|\frac{x+\frac1x-y-\frac1y}{x+y-\frac1x-\frac1y}\right|=\left|\frac{x-y}{x+y}\right|=\sqrt{1-\frac{4xy}{(x+y)^2}}=\sqrt{1-\frac4{a_nb_n}}.$$But we have $(a_n+b_n)^2\ge 4a_nb_n\implies 1-\frac{4}{a_nb_n}\le 1-\frac{16}{(a_n+b_n)^2},$ so$$\frac{|a_{n+1}-b_{n+1}|}{|a_n-b_n|}\le \sqrt{1-\frac{16}{(a_n+b_n)^2}}=\sqrt{1-\frac{16}{(a_0+b_0)^2}}.$$Now $\textstyle\sqrt{1-\frac{16}{(a_0+b_0)^2}}$ is a constant less than $1$, so $|a_n-b_n|$ converges to zero. It's now easy to see that $a_n,b_n$ are convergent as well, in fact they both converge to $\tfrac{a+b}{2}$; so invoking $(\star )$ and continuity, $$f(a)+f(b)=2f\left(\frac{a+b}{2}\right)\;\forall a,b>0 \text{ so that }ab\ge 4.\quad (\spadesuit)$$Okay, that's almost Jensen but with a constraint. How do we turn that into omnipotent Jensen? By introducing new variables, of course!
Now take arbitrary $a,b>0$. Choose sufficiently large $c,d$ so that:
  • $cd\ge 4$,
  • $bc\ge 4$,
  • $ad\ge 4$,
  • $\frac{a+d}{2}\cdot\frac{b+c}{2}\ge 4$, and
  • $\frac{a+b}{2}\cdot \frac{c+d}{2}\ge 4.$
These conditions are easy to satisfy by taking huge $c,d$; in fact, many of these are redundant. So by applying $(\spadesuit )$ multiple times, we have
\begin{align*}f(a)+f(b)+2f\left(\frac{c+d}{2}\right)&=f(a)+f(b)+f(c)+f(d)\\ &=2f\left(\frac{a+d}{2}\right)+2f\left(\frac{b+c}{2}\right)\\ &=4f\left(\frac{a+b+c+d}{2}\right)\\ &=2f\left(\frac{a+b}{2}\right)+2f\left(\frac{c+d}{2}\right).\end{align*}Therefore $f(a)+f(b)=2f\left(\tfrac{a+b}{2}\right)$ for arbitrary $a,b>0,$ so Jensen says $f(x)=cx+d$ for $c\ge 0$ (or just $c\in\mathbb R$ if the codomain is supposed to be $\mathbb R$, thanks @anantmudgal09 for pointing this out), which works nicely. $\blacksquare$

Chemical Analysis #4: Size matters
Quote:
(IMO 2011 P5) Let $f$ be a function from the set of integers to the set of positive integers. Suppose that, for any two integers $m$ and $n$, the difference $f(m) - f(n)$ is divisible by $f(m- n)$. Prove that, for all integers $m$ and $n$ with $f(m) \leq f(n)$, the number $f(n)$ is divisible by $f(m)$.

Proposed by Mahyar Sefidgaran, Iran

This may be IMO Problem 5 or a shortlist N5 or whatever, but I find this to be a massive troll. Not quite fitting of an Iranian problem. Tsk tsk.

Hmm, so looks like the solutions are $f(x)=x$...hey, wait, the range is positive numbers?! Where do negative numbers map to? Darn, $f$ is most likely some stupid function like constant or some ugly piecewise beast.

So let $P(m,n)$ denote the statement $f(m-n)|f(m)-f(n)$. As always, we start with plugging in nice things like $0,-m$ and stuff. So this gives: \begin{align*}P(m,0)&\implies f(m)|f(m)-f(0)\implies f(m)|f(0)\\P(0,m)&\implies f(-m)|f(0)-f(m)\implies f(-m)|f(m) \text{ since }f(-m)|f(0)\\ P(0,-m)&\implies f(m)|f(-m)\\ \implies & f(m)=f(-m)\end{align*}Wow, $f$ is even. Also, every $f(m)$ is a divisor of $f(0)$, so $f$ can take at most finitely many values. Can it get any sillier?

We want to finish off this thing as quickly as possible, so let's get to two variables. Choose $m,n$ and let's do whatever we can with them.
\begin{align*}P(m+n,m)&\implies f(n)|f(m+n)-f(m)\\ P(m+n,n)&\implies f(m)|f(m+n)-f(n)\\ P(m,-n)&\implies f(m+n)|f(m)-f(n)\end{align*}
Now we need a quick bookkeeping-ish job based on size-reasons. Suppose WLOG $f(m)> f(n)$, then $f(m+n)\le f(m)-f(n)<f(m)$, so $f(m)>|f(n)-f(m+n)|$, but we have $f(m)|f(n)-f(m+n)$, so $f(n)=f(m+n)$. Then $f(n)|f(m+n)|f(m)-f(n)\implies f(n)|f(m)$ follows immediately, proving the claim. $\blacksquare$

I hate it when the only non-trivial idea in a number theory problem is that divisibility implies size constraints. Fortunately, there's a better NT problem coming up to ease off that pang...

Chemical Analysis #5: Fair and square
Quote:
(Tournament of Towns 2012) Let $n$ be a positive integer. Prove that there exist integers $a_1, a_2, \cdots , a_n$ such that for any integer $x$, the number $(\cdots (((x^2+a_1)^2+a_2)^2+\cdots)^2+a_{n-1})^2+a_n$ is divisible by $2n-1$.

Lemme tell you a little secret: every ToT problem is discreetly a combinatorics problem. Trust me. Things may look disguised as something else, they might contain bits of non-trivial occurrences of divisibility, triangle, polynomials and what not, but deep down these are really attempts at masquerading. It's combinatorics all the way down.

In this problem, too, the key is combinatorial thinking. $x$ can take all $2n-1$ possible residues; we somehow need to restrict that to $0$ by squaring and adding stuff. Squaring $x$ once makes the number of residues bog down to $n$ already since $i^2\equiv (2n-1-i)^2\pmod{2n-1}$, so we're in good shape. Now we can add $n$ $a_i$'s and square lots of times in the meantime. It'd be really nice if adding $a_i$ and squaring reduced the number of possible residues everytime by at least $1$. In that case, we could decrease the possibilities by $1$ each until adding $a_{n-1}$ and squaring leaves us with only one possibility, in which case we could use a suitable $a_n$ to the whole thing to make that possibility zero.

Okay, let's see what we can do. Well, the only time the number of possible residues changes is when we square. But that collapses $i$ and $2n-1-i$ together. So maybe we can shift any two possible residues by $a_i$ at any given instant so that they become of the form $i,2n-1-i$, and then square? Well, let's say $a,b$ are two possible residues at some point. We want to find $a_i$ so that $(a+a_i)+(b+a_i)\equiv 2n-1\iff a_i\equiv\tfrac{2n-1-a-b}{2}\pmod{2n-1}$, and of course, this works, because $(2,2n-1)=1$, so division by $2$ makes sense! So we can add this $a_i$ and then square, so that the number of possible residues decreases by at least $1$, and proceed as described before. Mission accomplished! $\blacksquare$

Also, a fun fact: $\text{Fe}_3\text{N}_2$ is grey in colour, so you may slyly call it 'grey matter' and feel smug for the rest of the day.
This post has been edited 6 times. Last edited by Ankoganit, Jun 30, 2017, 3:59 AM
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    (i came here from searching google’s “nag a ram” and somehow ended up in aops again)

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  • Oh the blog is dead again.Fun fact nobody has even shouted since it's last post.

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  • monthly check finally sees its revival

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  • woohoo!! revive

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  • yay!!!!!

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  • @all proponents of revival: there you go.

    by Ankoganit, Jul 30, 2021, 8:23 AM

  • \revive \revive \revive

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  • Wow, this blog is amazinggg! Revive!!

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  • 'Elegant, not Elephant', 'Some properties of ferrous nitride'. Haha, creative titles I must say :). This is the reason I guessed you were one of the proposers of INMO P6 after reading 'A ninth-graders guide to polynomials.' Really love your work. :D

    by RMOAspirantFaraz, Mar 8, 2021, 1:20 PM

  • ;Puffer13

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  • @Prabh1512 the secret is lots of caffeine

    by Ankoganit, Sep 30, 2020, 6:12 AM

  • How to get the motivation to carry out random polynomial multiplication ? TO BASH of course

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