Found problems: 700
Four positive integers $x,y,z$ and $t$ satisfy the relations
\[ xy - zt = x + y = z + t. \]
Is it possible that both $xy$ and $zt$ are perfect squares?
Four positive integers $x,y,z$ and $t$ satisfy the relations
\[ xy - zt = x + y = z + t. \]
Is it possible that both $xy$ and $zt$ are perfect squares?
$(SWE 6)$ Prove that there are infinitely many positive integers that cannot be expressed as the sum of squares of three positive integers.
Solve the given equation in prime numbers $$p^3+q^3+r^3=p^2qr$$
Let $x_n$ the sequence defined by any nonnegatine integer $x_0$ and $x_{n+1}=1+\prod_{0 \leq i \leq n}{x_i}$
Show that there exists prime $p$ such that $p\not|x_n$ for any $n$.
Find all non negative integers $a{}$ and $b{}$ satisfying $5^a+6=31^b$
[i]Proposed by Erik Füredi, Budapest[/i]
Find all triples $(a,b,c)$ of integers satisfying $a^2 + b^2 + c^2 = 20122012$.
Determine all integer triplets $(x,y,z)$ such that
\[x^4+x^2=7^zy^2\mbox{.}\]
Denote $S$ as the set of prime divisors of all integers of form $2^{n^2+1} - 3^n, n \in Z^+$.
Prove that $S$ and $P-S$ both contain infinitely many elements (where $P$ is set of prime numbers).
Let $d$ be any positive integer not equal to 2, 5, or 13. Show that one can find distinct $a$ and $b$ in the set $\{2,5,13,d\}$ such that $ab - 1$ is not a perfect square.
Let $p$ be an odd prime number and $M$ a set derived from $\frac{p^2 + 1}{2}$ square numbers.
Investigate whether $p$ elements can be selected from this set whose arithmetic mean is an integer.
(Walther Janous)
Let $n$ be a positive integer. Show that there are infinitely many primes $p$ such that the smallest positive primitive root of $p$ is greater than $n$.
Let $p$ be a polynomial with integer coefficients and let $a_1<a_2<\cdots <a_k$ be integers. Given that $p(a_i)\ne 0\forall\; i=1,2,\cdots, k$.
[list]
(a) Prove $\exists\; a\in \mathbb{Z}$ such that
\[ p(a_i)\mid p(a)\;\;\forall i=1,2,\dots ,k \]
(b) Does there exist $a\in \mathbb{Z}$ such that
\[ \prod_{i=1}^{k}p(a_i)\mid p(a) \][/list]
Positive integers $x>1$ and $y$ satisfy an equation $2x^2-1=y^{15}$. Prove that 5 divides $x$.
Prove that there exist infinitely many positive integers $n$ such that the largest prime divisor of $n^4 + n^2 + 1$ is equal to the largest prime divisor of $(n+1)^4 + (n+1)^2 +1$.
Let $p$ a prime number and $r$ an integer such that $p|r^7-1$. Prove that if there exist integers $a, b$ such that $p|r+1-a^2$ and $p|r^2+1-b^2$, then there exist an integer $c$ such that $p|r^3+1-c^2$.
Prove that if $ p$ is a prime number, then $ 7p+3^{p}-4$ is not a perfect square.
The integers $a$ and $b$ have the property that for every nonnegative integer $n$ the number of $2^n{a}+b$ is the square of an integer. Show that $a=0$.
Find the smallest natural number $ k$ for which there exists natural numbers $ m$ and $ n$ such that $ 1324 \plus{} 279m \plus{} 5^n$ is $ k$-th power of some natural number.
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]
Let $a$ and $b$ be integers. Prove that $\frac{2a^2-1}{b^2+2}$ is not an integer.
Let $ p \geq 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1,2,\ldots, p-1 \}$ that was not chosen before by either of the two players and then chooses an element $a_i$ from the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices have been chosen .Then the following number is computed:
$$M=a_0+a_110+a_210^2+\cdots+a_{p-1}10^{p-1}= \sum_{i=0}^{p-1}a_i.10^i$$.
The goal of Eduardo is to make $M$ divisible by $p$, and the goal of Fernando is to prevent this.
Prove that Eduardo has a winning strategy.
[i]Proposed by Amine Natik, Morocco[/i]
Solve the equation
\[ 3^x \minus{} 5^y \equal{} z^2.\]
in positive integers.
[i]Greece[/i]
Let $z$ be a positive integer that is not divisible by $8$. Furthermore, let $n \geqslant 2$ be a positive integer. Prove that none of the numbers of the form $z^n + z + 1$ is a square number.
[i](Walther Janous)[/i]
Find all prime number $p$ such that the number of positive integer pair $(x,y)$ satisfy the following is not $29$.
[list]
[*]$1\le x,y\le 29$
[*]$29\mid y^2-x^p-26$
[/list]