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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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jlacosta
Mar 2, 2025
0 replies
Perfect Squares, Infinite Integers and Integers
steven_zhang123   2
N 12 minutes ago by ohiorizzler1434
Source: China TST 2001 Quiz 5 P1
For which integer \( h \), are there infinitely many positive integers \( n \) such that \( \lfloor \sqrt{h^2 + 1} \cdot n \rfloor \) is a perfect square? (Here \( \lfloor x \rfloor \) denotes the integer part of the real number \( x \)?
2 replies
+2 w
steven_zhang123
Yesterday at 12:06 PM
ohiorizzler1434
12 minutes ago
Line Perpendicular to Euler Line
tastymath75025   55
N 15 minutes ago by ohiorizzler1434
Source: USA TSTST 2017 Problem 1, by Ray Li
Let $ABC$ be a triangle with circumcircle $\Gamma$, circumcenter $O$, and orthocenter $H$. Assume that $AB\neq AC$ and that $\angle A \neq 90^{\circ}$. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively, and let $E$ and $F$ be the feet of the altitudes from $B$ and $C$ in $\triangle ABC$, respectively. Let $P$ be the intersection of line $MN$ with the tangent line to $\Gamma$ at $A$. Let $Q$ be the intersection point, other than $A$, of $\Gamma$ with the circumcircle of $\triangle AEF$. Let $R$ be the intersection of lines $AQ$ and $EF$. Prove that $PR\perp OH$.

Proposed by Ray Li
55 replies
tastymath75025
Jun 29, 2017
ohiorizzler1434
15 minutes ago
Foot from vertex to Euler line
cjquines0   31
N 19 minutes ago by pUssydestroyer777
Source: 2016 IMO Shortlist G5
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $ABC$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $AB$ and $AC$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $BC$, and let $M$ be the midpoint of $BC$. Prove that the circumcentre of triangle $XSY$ is equidistant from $P$ and $M$.
31 replies
+1 w
cjquines0
Jul 19, 2017
pUssydestroyer777
19 minutes ago
Inequality => square
Rushil   12
N an hour ago by ohiorizzler1434
Source: INMO 1998 Problem 4
Suppose $ABCD$ is a cyclic quadrilateral inscribed in a circle of radius one unit. If $AB \cdot BC \cdot CD \cdot DA \geq 4$, prove that $ABCD$ is a square.
12 replies
Rushil
Oct 7, 2005
ohiorizzler1434
an hour ago
No more topics!
Another NT FE
nukelauncher   59
N Yesterday at 6:17 PM by AshAuktober
Source: ISL 2019 N4
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
59 replies
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nukelauncher
Sep 22, 2020
AshAuktober
Yesterday at 6:17 PM
Another NT FE
G H J
Source: ISL 2019 N4
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nukelauncher
354 posts
#1 • 8 Y
Y by TheLiker, itslumi, Kanep, centslordm, megarnie, ImSh95, TheHU-1729, deplasmanyollari
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
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nukelauncher
354 posts
#2 • 3 Y
Y by Mathematicsislovely, centslordm, ImSh95
Solution
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anyone__42
92 posts
#3 • 11 Y
Y by idkwhatthisis, Kanep, Aryan-23, centslordm, TheMath_boy, uvuvwevwevwe, megarnie, sabkx, Vladimir_Djurica, ImSh95, TensorGuy666
can be done easily using dirichlet's theorem
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dangerousliri
924 posts
#4 • 4 Y
Y by A-Thought-Of-God, centslordm, sabkx, ImSh95
Here it is a solution that I did find when I did get the shortlist. Somehow for me it took me faster to solve this then N1 which was on the test. Also it should be $a+b>C$ and not $a+b>2019.$

Since $gcd(f(1),1)=1$ from Dirichlet's theorem we have there exist infinitely prime numbers of the form $1+nf(1)$. Taking all such $n>C$ then for $a=1$ and $b=n$ we have,
$$1+f(n)|1+nf(1)\Rightarrow f(n)=nf(1).$$
Now taking $a$ fixed and $b=n>C$ large enough we have,
$$a+nf(1)|a+f(n)|a^2+nf(a)|f(1)a^2+f(1)nf(a)\Rightarrow a+nf(1)|f(1)a^2+f(1)nf(a)-(a+nf(1))f(a)$$$$\Rightarrow a+nf(1)|a(f(1)a-f(a))\Rightarrow f(a)=f(1)a=ka.$$
Hence $f(a)=ka$ for all positive integers $a$, and clearly for any positive integer $k$ hold.
This post has been edited 1 time. Last edited by dangerousliri, Sep 23, 2020, 12:10 AM
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blacksheep2003
1081 posts
#5 • 3 Y
Y by centslordm, sabkx, ImSh95
Solution

Oh wow, I thought my solution was pretty unique to use Dirichlet's Theorem, but it seems other people had the same idea lol oops :P
This post has been edited 4 times. Last edited by blacksheep2003, Sep 23, 2020, 12:49 AM
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MarkBcc168
1593 posts
#6 • 12 Y
Y by mijail, Imayormaynotknowcalculus, k12byda5h, A-Thought-Of-God, Aryan-23, Kanep, centslordm, tigerzhang, GeoKing, ImSh95, Stuffybear, iamnotgentle
Comment: Extremely standard NT functional divisibility. This trick used to solve this problem have been appeared many many times.

As in every functional divisibility problems, the main point is to prove that $f$ is linear on some infinite set. We prove the following claim.

Claim: There exists infinite sequence $C<x_1<x_2<\hdots$ such that $f(x_i)=kx_i$ for some constant $k$.

Proof: The idea is to force the right hand side to be prime. By Dirichlet, there exists infinitely many $x>C$ which $1+xf(1)=p$ is prime. However,
$$1+f(x)\mid xf(1)+1\implies f(x)+1\in\{1,p\}\implies f(x)=xf(1)$$for infinitely many $x$. Hence we are done. $\blacksquare$

Once we have this we are almost done. Fix $n\in\mathbb{Z}^+$. Since
\begin{align*}
&n+kx_i\mid n^2+x_if(n) \\ 
\implies &n+kx_i\mid k(n^2+x_if(n)) - f(n)(n+kx_i) \\
\implies &n+kx_i\mid kn^2-nf(n)
\end{align*}and $x_i$ is arbitrarily large. We get $f(n)=kn$ for any $n$.
This post has been edited 1 time. Last edited by MarkBcc168, Sep 23, 2020, 1:43 AM
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FAA2533
65 posts
#7 • 3 Y
Y by Aryan-23, centslordm, ImSh95
nukelauncher wrote:
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>2019$.
It somehow seemed very easy to me than N1.

Solution
This post has been edited 6 times. Last edited by FAA2533, Jul 18, 2022, 2:35 PM
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ubermensch
820 posts
#9 • 4 Y
Y by ATGY, Illuzion, centslordm, ImSh95
Let $P(a,b)$ denote the assertion that $a+f(b) \mid a^2+bf(a)$.

Naturally, a $P(1,b)$ substitution gives us $1+f(b) \mid 1+bf(1) \implies bf(1) \geq f(b)$. We'll now denote $f(1)$ by $k$, as we shall be using it quite extensively.

$P(a,1): a+k \mid a^2+f(a) \implies a+k \mid a^2+f(a)-a(a+k) \implies a+k \mid f(a)-ak \implies a+k \mid f(a)-ak +k(a+k)$.

Hence we get $a+k \mid f(a)+k^2$. Coupled with our first bound, this gives us that $f(a)=at+tk-k^2$ for large enough $a$.

Here, although it is quite instinctual to substitute $a$ as a prime, we can in fact avoid that altogether and deal just with our regular naturals- although it will get quite ugly, but at the very least it will remain very mechanical.

Our aim, as it always is in such problems, is to manipulate the divisibility so as to be able to make the left hand side arbitrarily large without altering the expression on the right, hence forcing the right hand side to be become zero. A simple $(a,b)$ substitution in the original expression with a little modification would give us what we're looking for.

Firstly-
$$a+f(b) \mid a^2+bf(a) \implies a+f(b) \mid a^2+bf(a)-a(a+f(b))$$$$\implies a+f(b) \mid bf(a)-af(b) +f(b)(a+f(b)) \implies a+f(b) \mid bf(a)+f(b)^2$$
Thus we have $a+f(b) \mid f(b)^2 + bf(a)$. We now substitute $f(a)=at+(k^2-kt)$ to get-

$$a+f(b) \mid b(at+k^2-kt)+f(b)^2$$$$\implies a+f(b) \mid abt+bk^2-bkt+f(b)^2-bt(a+f(b)) \implies a+f(b) \mid b(k^2-kt)+f(b)^2-btf(b)$$Observe now that we can make our $a$ arbitrarily large for a constant $b$, and noting that $k=f(1)$ has a fixed value along with $t \leq k$ forces $f(b)^2-btf(b)+b(k^2-kt)=0$, as planned.

Writing $f(b)=rb+rk-k^2$ gives us $(rb+rk-k^2)^2-bt(rb+rk-k^2)+b(k^2-kt)=0$. For large $b$, since our $k$ and $t$ are small (compared to our now large $b$), we can simply equate the coefficients of $b^2$ and $b$ and the constant term zero individually. Hence, $r^2-rt=0 \implies r=t$ and equating the constant term to zero gives us $r=k$.

Thus, for arbitrarily large $n$, we have $f(n)=kn$. The finish is now quite simple-
$$P(n,b): n+f(b) \mid n^2+knb-n(n+f(b))-(kb-f(b))(n+f(b))$$$$\implies n+f(b) \mid (kb-f(b))f(b)$$for large $n$, hence forcing $f(b)=kb$ for all $b$.

Clearly $f(n)=kn$ satisfies the original condition, and hence we're done.

Quite a standard one, and rather mechanical, especially if you've done something like this before (POP). Nevertheless, quite an instructive exercise on functional divisibilities...
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dchenmathcounts
2443 posts
#10 • 3 Y
Y by centslordm, Marie36, ImSh95
guess this has been said before, but here goes

We claim the answer is $f(x)=kx$ for $k\in \mathbb{Z}_{>0}.$

Fix $a=1.$ Note that $1+bf(1)$ is a prime infinitely often by Dirichlet (is this how you spell it?). So let $b$ be some absurdly large number such that $a+b>2019$ and $1+bf(1)$ is prime; then $1+f(b)\mid 1+bf(1).$ Since $1+bf(1)$ is prime we either have $1+f(b)=1$ or $1+f(b)=1+bf(1).$ Clearly the latter holds, so there are infinitely many $b$ such that $f(b)=bf(1).$

Now take some mega ultra large $b$ such that $f(b)=bf(1),$ and note that $a+f(b)=a+bf(1)\mid a^2+bf(a)$ and $a+bf(1)\mid a^2+abf(1).$ Subtracting the latter from the former gives $a+bf(1)\mid b(f(a)-af(1)).$ Note that $\gcd(a+bf(1),b)=\gcd(a,b),$ so if $a$ is prime this implies that $a+bf(1)\mid f(a)-af(1).$ Since $b$ is, as I quote, "ultra mega large," $|a+bf(1)|>|f(a)-af(1)|,$ so for all prime $a$ we have $f(a)=af(1).$

Now take some ultra mega large prime $b,$ which we have just shown exists, and note that
\[a+bf(1)\mid b(f(a)-af(1)), \text{ which implies that}\]\[a+bf(1)\mid f(a)-af(1),\]and since $b$ is ultra mega large, we have shown that $f(a)=af(1)$ in general.
This post has been edited 1 time. Last edited by dchenmathcounts, Sep 23, 2020, 5:14 PM
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ayan_mathematics_king
1527 posts
#11 • 6 Y
Y by DapperPeppermint, Mathematicsislovely, amar_04, A-Thought-Of-God, centslordm, ImSh95
Solution by my friend RustyFox who is banned now.So I am posting his solution.
nukelauncher wrote:
Find all functions $f:\mathbb Z_{>0}\to \mathbb Z_{>0}$ such that $a+f(b)$ divides $a^2+bf(a)$ for all positive integers $a$ and $b$ with $a+b>C$.

Solution : Let $P(x, y)$ be the assertion. $P(1, k) \implies kf(1)\geq f(k)$.

This is going to be useful. Let $p \mid x$ where $p$ is a prime.

Let $k = Cf(x)$

$P(kp-f(x), x \implies kp \mid (kp-f(x))^2 + f(kp-f(x))x$ which means that $p \mid f(x)$. So, $p \mid f(p)$ for all primes $p$.

Let us say that $f(p)= g(p)p$ for primes $p$. Observe that $0 < g(p) \leq f(1)$. If $g(p)\neq g(q)$ for two distinct primes $p, q$ which are much much greater than $C$ (say it's much greater than $(C+2019f(1))!\uparrow \uparrow (C+2019f(1))!)$, then $P(p, q) \implies p + qg(q) \mid p^2 + pqg(p)$ or in other words $p + qg(q) \mid pq(g(p)-g(q))$. Now since $gcd(p, q) = gcd(p, g(q)) = gcd(p, qg(q)) = 1$, we havs $p + qg(q) \nmid pq$. Now, $p + qg(q) \mid g(p)-g(q)$ but the dividend is less than $f(1)$ and due to our value of $p, q$, $p + qg(q) >> f(1)$ which means that for all sufficiently large primes we have $g(p) = g(q) = c$. Now for any prime $p$ such that $f(p) = pc$ and any other prime $q$ for which we aren't sure if $f(q)$ equals $qc$ or not and obviously here $q < p$,we take $P(q, p)$ to get that $q + pc \mid p(f(q)-qc)$ and since $pc  + q > pc \geq p$, we havs $q + pc \mid f(q)-qc$ and so see that $f(q) - qc \leq qf(1)-qc = q(f(1)-c)$ and so choosing $p > q(f(1)-c)$ would suffice and so for all primes $p$, we have $f(p)=pc$. Now, let $p >>a$. Then $P(a, p) \implies a + pc \mid pf(a)-pca$ but $gcd(p, a)=1$ and so $a+pc \mid f(a)-ac$. Choosing large enough $p$ yields that $\boxed{f(a)=ac}$ for all naturals $a$
This post has been edited 3 times. Last edited by ayan_mathematics_king, Sep 24, 2020, 7:51 AM
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ProblemSolver2048
104 posts
#15 • 2 Y
Y by centslordm, ImSh95
Since gcd(f(1,),1)=1 from Dirichlet's theorem we have there infinitely prime numbers of the form 1+nf(1). Taking all such n>C then for all a=1 and b=n we have $$1+f(n)|1+nf(1)\Rightarrow f(n)=nf(1).$$Now taking a fixed and b=n> C large enough we have,
$$a+nf(1)|a+f(n)|a^2+nf(a)|f(1)a^2+f(1)nf(a)\Rightarrow a+nf(1)|f(1)a^2+f(1)nf(a)-(a+nf(1))f(a)$$$$\Rightarrow a+nf(1)|a(f(1)a-f(a))\Rightarrow f(a)=f(1)a=ka.$$hence f(a)=ka for all positibe integers , clearly for any positive integer k hold.
Acknowledgements to dangerousliri's in post 4. I didn't make it.
This post has been edited 1 time. Last edited by ProblemSolver2048, Dec 14, 2020, 7:39 PM
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Mathematicsislovely
245 posts
#16 • 5 Y
Y by amar_04, centslordm, ImSh95, sabkx, ylmath123
Let $P(a,b)$ denote the assertion of the question.

Claim:For all prime $p$ we have $p|f(p)$.
proof.Fix a prime $p$.$k$ be an integer such that $(k+1)p-f(p)>2020^{2020}$.Then $P(kp-f(p),p)\implies p|kp-f(p)+f(p)|(kp-f(p))^2+pf(stuff)\implies p|f(p)$. $\blacksquare$.

Then $f(p_n)=c_np_n$ where $p_n$ is the $n^{th}$ prime.But as $P(1,b)\implies f(b)\le f(1)b$ $\forall 
 b\ge 2020$. So $c_n$ is bounded sequence.Hence,there exists an integer $r$ such that it is equal to infinitely many $c_i$.So there exists an increasing sequence $(q_n)_{n=1}^{\infty}$ of prime integers such that $f(q_s)=rq_s$ $\forall s\in \mathbb N$.

Now, for a fix $a$ take a big $q_i$ such that $gcd(a,q_i)=1$.Then,
$P(a,q_i)\implies\\
 a+rq_i|a^2+q_if(a)\\
\implies a+rq_i|a^2+q_if(a)-a^2-arq_i\\
\implies a+rq_i|q_i(f(a)-ar)\\
\implies a+rq_i|(f(a)-ar)$.
Now for sufficiently large $q_i$ we get $f(a)=ar$.Hence,$\boxed{f(x)=rx}$ is the only solution which indeed satisfies the functional equation. $\blacksquare$.
This post has been edited 4 times. Last edited by Mathematicsislovely, Sep 26, 2020, 11:00 PM
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Ilove_mathematics
188 posts
#17 • 3 Y
Y by centslordm, odkaI, ImSh95
Let $P(a,b)$ denote the divisibility $(a+f(b)) \mid (a^2+bf(a)).  ($For simplicity say $C = 2019)$

Let $n$ denote $f(1).$ We are going to show that $f(k) = kn,$ for all $k \in \mathbb{Z}_{>0}.$ Let $\mathbb{A} = \{C+1, C+2, ...\}.$

Lemma 1: Define $g:\mathbb{A} \rightarrow \mathbb{Q}$ as $g(a) = \frac{an-f(a)}{a+n},$ then $g$ is limited.

Proof: Let $a$ be a positive integer greater than $C.$ Then $P(a,1)$ gives $a+n \mid a^2 + f(a) \implies (a+n).q = a^2 + f(a)$ with $q \in \mathbb{N}.$ On the other hand $P(1, a) \implies 1+f(a) \mid 1+an \implies f(a) \leq an,$ therefore $(a+n).a =$ $a^2 + an \geq a^2 + f(a), $ suppose for the sake of contradiction that $f(a) < an \implies q < a \implies q = a-k$ so $(a+n)(a-k) = a^2 + f(a) > a^2,$ since $(a+n)(a-n)  < a^2$ we have $k < n \implies k$ is bounded as $a$ varies along the natural numbers greater than $C,$ yielding $k = \frac{an-f(a)}{a+n} = g(a)$ is limited.$\blacksquare$

Lemma 2: There are infinitely many positive integers $s$ such that $f(s) = sn.$

Proof: Let $S = \{a_1, a_2, ...\}$ be the set of positive integers such that $\frac{a_in-f(a_i)}{a_i+n} = t$ a constant, with $C < a_1 < a_2 < ...,$ there exists such $t,$ because the set $\mathbb{A}$ is unbounded. So $\frac{a_in-f(a_i)}{a_i+n} = t \implies$ $f(a_i) = a_in-t(a_i+n),$ for all $i.$
Now fix $j$ such that $a_j(n-t+1) \ne nt.$ Plugging $a = a_i, b = a_j$ gives: $$(a_i + f(a_j)) \mid (a_i^2 + a_jf(a_i)) \implies$$$$(a_i + a_jn - t(a_j+n)) \mid [(a_i^2 + a_ja_in - a_jt(a_i+n)) - a_i(a_i+a_jn-t(a_j+n))] \implies$$$$(a_i+a_jn-t(a_j+n)) \mid (a_i-a_j)nt.$$Let $d_i$ be $gcd(a_i + a_jn-t(a_j+n), a_i - a_j),$ (of course $i \ne j$) if $d_i > max(a_j(n-t+1), nt) \implies$ $a_i + a_jn-t(a_j+n) \equiv a_j + a_j(n-t) - nt \pmod {d_i} \implies d_i \mid a_j(n-t+1)-nt,$ as supposed we must have $a_j(n-t+1) = nt,$ contradiction. Therefore $d_i$ is bounded. Observe $(a_i + a_jn-t(a_j+n))|d_int,$ $\forall i \in \mathbb{N}, i \ne j,$ setting $i \rightarrow \infty \implies a_i \rightarrow \infty$ thus, we eventually have: $a_i + a_jn -t(a_j+n) > d_int \implies d_int = 0 \implies t = 0 \implies f(a_i) = a_in,$ for all $i,$ sufficiently large. $\blacksquare$


Fix $k$ a positive integer.
Plug $a = k, b = a_i \implies k+a_in \mid k^2 + a_if(k) \implies (k+a_in) \mid (k^2+a_if(k) - k(k+a_in)),$ yielding $(k+a_in) \mid a_i(f(k)-kn)$ so $(k+a_in).q_i = a_i.|f(k)-kn|, q_i \in \mathbb{Z}_{\geq 0} \implies q_i = \frac{a_i.|f(k)-kn|}{k+a_in}$ call $t = |f(k)-kn|$ a constant. If $q_i > t \implies (k+a_in).t < a_i.t$ contradiction. We conclude $q_i$ is bounded. On the other hand if $i \ne j$ and $q_i = q_j \implies \frac{a_it}{k+a_in} = \frac{a_jt}{k+a_jn}$ $\implies a_itk + a_ia_jnt = a_jtk + a_ja_int \implies (a_i-a_j)kt = 0$ then $t$ would be $0.$ Since $q_i, q_j$ are bounded, both positive and we can choose $i, j \in \mathbb{N}$ such that $q_i = q_j, f(k)-kn = 0$ for all $k \in \mathbb{N},$ as $(a+kb).a = a^2 + akb \implies f(a) = an$ for all positive integers $n.$
This post has been edited 1 time. Last edited by Ilove_mathematics, Sep 27, 2020, 4:22 AM
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solver1104
510 posts
#18 • 2 Y
Y by centslordm, ImSh95
Remark: Standard problem, seems pretty easy in difficulty unless I messed up. Also, I cite the so called Weak - Dirichlet theorem many times over this proof. Although no proof of Dirichlet Theorem that I know of is elementary, Weak Dirichlet can easily be proven by cyclotomic polynomials - see Evan Chen's Orders handout for more details.

random stuff

We claim that the answer is all functions $f(x) \equiv cx$ for any positive integer $c$. It's easy to see that all such functions work, because $a + bc | a^2 + abc = a(a+bc)$. Now we show that these are the only functions that work.

Consider the more general problem, replacing $2019$ with any constant $C$. Denote the given assertion$P(a,b)$.

Now we claim that $f(b) = bf(1)$ holds for infinitely many $b$. Indeed, $P(1,b)$ gives $1 + f(b) | 1 + bf(1)$, and now by Weak Dirichlet, there exists infinitely many $b$ such that $bf(1) + 1$ is prime, and since $1 + f(b) > 1$, we have $f(b) = bf(1)$, as desired.

Now we show that this actually implies $f(a) = af(1)$ for all $a$. Indeed, by Weak Dirichlet again, there exists infinitely many primes that are $1 \pmod{f(a) \cdot f(1)}$ for any $a$. Now take a prime of this form, $p$, and let $p$ be arbitrarily large. Note that $p$ is relatively prime to $f(a)$, and it is also of the form $cf(1) + 1$ for some $c$, hence by the first claim, $f(p) = pf(1)$. Then $P(p,a)$ gives $p + f(a) | p^2 + apf(1)$. Now $\gcd(p, p+f(a)) = \gcd(p,f(a)) = 1$ by the assumption that $p$ is $1 \pmod{f(a)}$, so the extra factor of $p$ on the $RHS$ can be taken out. Now we have $p + f(a) | p + af(1)$ for all $a$, and since $p$ is arbitrarily large, so that $p >>> f(a), af(1)$, so the two sides are actually equal, and we are done.
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rcorreaa
238 posts
#19 • 2 Y
Y by centslordm, ImSh95
Pretty standard NT FE problem... :maybe:

From the problem, $1+f(n)|1+nf(1)$ for all $n \in \mathbb{N}$. From Dirichlet's Theorem, since $gcd(1,f(1))=1$, there are infinetly prime numbers in the AP $\{1+nf(1)\}_{n \geq 1}$.

Thus, since $1+f(n)>1, 1+f(n)=1+nf(1)$ when $1+nf(1)$ is prime $\implies f(n)=nf(1)$ in this case.

Hence, there exists a sequence $\{a_n\}_{n=1}^{\infty}$ such that $f(a_n)=a_nf(1)$ for all $n \in \mathbb{N}$. Then, from the problem,
$$a+a_nf(1)|a^2+a_nf(a) \implies a+a_nf(1)|a^2f(1)+a_nf(a)f(1),af(a)+f(a)f(1)a_n$$$$\implies a+a_nf(1)|a^2f(1)-af(a)$$for all $n \in \mathbb{N}$

Taking $n$ large enough, $a^2f(1)-af(a)$ must be $0$.

Therefore, $f(a)=af(1)$ for all $a \in \mathbb{N}$, which clearly works.

$\blacksquare$
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