Found problems: 700
Find all $n,p,q\in \mathbb N$ that:\[2^n+n^2=3^p7^q\]
Consider a sequence $T_0, T_1, \dots$ of polynomials defined recursively by $T_0(x) = 2$, $T_1(x)=x$, and $T_{n+2}(x) = xT_{n+1}(x) - T_n(x)$ for each nonnegative integer $n$. Let $L_n$ be the sequence of Lucas Numbers, defined by $L_0 = 2$, $L_1 = 1$, and $L_{n+2} = L_n+L_{n+1}$ for every nonnegative integer $n$.
Find the remainder when $ T_0\left( L_0 \right) + T_1 \left( L_2 \right) + T_2 \left( L_4 \right) + \dots + T_{359} \left( L_{718} \right)$ is divided by $359$.
[i]Proposed by Yang Liu[/i]
Let $k \in \mathbb{Z}^+$ and set $n=2^k+1.$ Prove that $n$ is a prime number if and only if the following holds: there is a permutation $a_{1},\ldots,a_{n-1}$ of the numbers $1,2, \ldots, n-1$ and a sequence of integers $g_{1},\ldots,g_{n-1},$ such that $n$ divides $g^{a_i}_i - a_{i+1}$ for every $i \in \{1,2,\ldots,n-1\},$ where we set $a_n = a_1.$
[i]Proposed by Vasily Astakhov, Russia[/i]
Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square.
For every positive integer $n$, let $S_n$ be the sum of the first $n$ prime numbers:
$S_1 = 2, S_2 = 2 + 3 = 5, S_3 = 2 + 3 + 5 = 10$, etc. Can both $S_n$ and $S_{n+1}$
be perfect squares?
Decide whether or not there exists a nonconstant polynomial $Q(x)$ with integer coefficients with the following property: for every positive integer $n > 2$, the numbers \[ Q(0), \; Q(1), Q(2), \; \dots, \; Q(n-1) \] produce at most $0.499n$ distinct residues when taken modulo $n$.
[i]Proposed by Yang Liu[/i]
Consider the set $S = \{ q + \frac{1}{q}, \text{ where } q \text{ ranges over all positive rational numbers} \}$.
(a) Let $N$ be a positive integer. Show that $N$ is the sum of two elements of $S$ if and only if $N$ is the product of two elements of $S$.
(b) Show that there exist infinitely many positive integers $N$ that cannot be written as the sum of two elements of $S$.
(c)Show that there exist infinitely many positive integers $N$ that can be written as the sum of two elements of $S$.
Determine all integers $n \ge 3$ whose decimal expansion has less than $20$ digits, such that every quadratic non-residue modulo $n$ is a primitive root modulo $n$.
[i]An integer $a$ is a quadratic non-residue modulo $n$, if there is no integer $b$ such that $a - b^2$ is divisible by $n$. An integer $a$ is a primitive root modulo $n$, if for every integer $b$ relatively prime to n there is a positive integer $k$ such that $a^k - b$ is divisible by $n$.[/i]
Find all pairs $(p, q)$ of prime numbers satisfying
\[ p^3+7q=q^9+5p^2+18p. \]
Prove that there exists infinitely many positive integers $n$ for which the equation$$x^2+y^{11}-z^{2022!}=n$$has no solution $(x,y,z)$ over the integers.
Form the infinite graph $A$ by taking the set of primes $p$ congruent to $1\pmod{4}$, and connecting $p$ and $q$ if they are quadratic residues modulo each other. Do the same for a graph $B$ with the primes $1\pmod{8}$. Show $A$ and $B$ are isomorphic to each other.
[i]Linus Hamilton.[/i]
Prove that if $n$ is a positive integer such that the equation \[ x^3-3xy^2+y^3=n \] has a solution in integers $x,y$, then it has at least three such solutions. Show that the equation has no solutions in integers for $n=2891$.
Given functions $f,g:N^*\rightarrow N^*$, $g$ is surjective and $2f(n)^2=n^2+g(n)^2$, $\forall n>0$. Prove that if $|f(n)-n|\le2005\sqrt n$, $\forall n>0$, then $f(n)=n$ for infinitely many $n$.
Show that there do not exist positive integers $m$ and $n$ such that \[ \dfrac{m}{n} + \dfrac{n+1}{m} = 4 . \]
Find all functions $g:\mathbb{N}\rightarrow\mathbb{N}$ such that \[\left(g(m)+n\right)\left(g(n)+m\right)\] is a perfect square for all $m,n\in\mathbb{N}.$
[i]Proposed by Gabriel Carroll, USA[/i]
[b](a)[/b] Determine if exist an integer $n$ such that $n^2 -k$ has exactly $10$ positive divisors for each $k = 1, 2, 3.$
[b](b) [/b]Show that the number of positive divisors of $n^2 -4$ is not $10$ for any integer $n.$
Does there exist a nonzero polynomial $P(x)$ with integer coefficients satisfying both of the following conditions?
[list]
[*]$P(x)$ has no rational root;
[*]For every positive integer $n$, there exists an integer $m$ such that $n$ divides $P(m)$.
[/list]
In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled.
[i]Proposed by Jerry Xu[/i]
[hide=Solution]
[i]Solution.[/i] $\boxed{21}$
Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows:
Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane.
[asy]
import geometry;
size(8cm);
pair A = (0,sqrt(3));
pair B = (-1,0);
pair C = (1,0);
pair M = (0,0);
for (int i = -1; i <= 2; ++i) {
draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3)));
draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3)));
draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3)));
}
draw(A--B--C--A, red);
dot(M);
label("$A$",A+(0,0.25),N);
label("$B$",B-(0.25,0),SW);
label("$C$",C+(0.25,0),SE);
label("$M$",M,S);
[/asy]
It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases:
$1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that
$$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case.
By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution
$$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$).
We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution).
From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are
\begin{align*}
(x,y)&=(0,1),(4,7),(56,97) \\
\longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97)
\end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex).
For $k=3$, using the multiplicative principle we get two new smaller solutions
\begin{align*}
(x,y)&=(0,3),(12,21) \\
\longrightarrow (n,m,l)&=(0,1,3),(12,1,21)
\end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$.
For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$.
Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$.
$2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that
$$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that
$$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions.
Our final answer is thus $1+7+13=\boxed{21}$.
[/hide]
Positive integers $x>1$ and $y$ satisfy an equation $2x^2-1=y^{15}$. Prove that 5 divides $x$.
Let $p>13$ be a prime of the form $2q+1$, where $q$ is prime. Find the number of ordered pairs of integers $(m,n)$ such that $0\le m<n<p-1$ and
\[3^m+(-12)^m\equiv 3^n+(-12)^n\pmod{p}.\]
[i]Alex Zhu.[/i]
[hide="Note"]The original version asked for the number of solutions to $2^m+3^m\equiv 2^n+3^n\pmod{p}$ (still $0\le m<n<p-1$), where $p$ is a Fermat prime.[/hide]
Let $d$ be any positive integer not equal to $2, 5$ or $13$. Show that one can find distinct $a,b$ in the set $\{2,5,13,d\}$ such that $ab-1$ is not a perfect square.
Find all functions $f:\mathbb{N}\to\mathbb{N}$ such that
$$|xf(y)-yf(x)|$$
is a perfect square for every $x,y \in \mathbb{N}$
Prove that among any $43$ positive integers there exist two $a$ and $b$ such that $a^2 - b^2$ is divisible by $100$.
Let $m$ and $n$ be natural numbers such that \[A=\frac{(m+3)^{n}+1}{3m}\] is an integer. Prove that $A$ is odd.
On a blackboard, there are $17$ integers not divisible by $17$. Alice and Bob play a game.
Alice starts and they alternately play the following moves:
$\bullet$ Alice chooses a number $a$ on the blackboard and replaces it with $a^2$
$\bullet$ Bob chooses a number $b$ on the blackboard and replaces it with $b^3$.
Alice wins if the sum of the numbers on the blackboard is a multiple of $17$ after a finite number of steps.
Prove that Alice has a winning strategy.
(Daniel Holmes)