Solve this equation in positive integers $$x^3 - y^3 = xy + 61$$
Solve in integers the equation $$(x+y)^2 = x^3 + y^3$$
Let $k$ be a positive integer, show that $(4k+3)$ cannot be a square number.
How many numbers in this series are squares? $$1, 14, 144, 1444, 14444, \cdots$$
Find all positive integer $n$ such that $n$ is a square and its last four digits are the same.
Solve the following equation in positive integers: $15x - 35y + 3 = z^2$
Find a four-digit square number whose first two digits are the same and the last two digits are the same too.
Solve the following equation in positive integers: $3\times (5x + 1)=y^2$
Find all pairs of integers $(x, y)$ such that $5\times (x^2 + 3)= y^2$.
If we arrange all the square numbers ascendingly as a queue: $1491625364964\cdots$ What is the $612^{th}$ digit?
In how many zeros does the number $\frac{2002!}{(1001!)^2}$ end?
What is the maximum value of $n$ for which there is a set of distinct positive integers $k_1$, $k_2$, $\dots$ , $k_n$ for which $$k_1^2 + k_2^2 + \cdots +k_n^2 = 2002$$
Under the new AMC 10, 12 scoring method, 6 points are given for each correct answer, 2.5 points are given for each unanswered question, and no points are given for an incorrect answer. Some of the possible scores between 0 and 150
can be obtained in only one way, for example, the only way to obtain a score of 146.5 is to have 24 correct answers and one unanswered question. Some scores can be obtained in exactly two ways; for example, a score of 104.5 can
be obtained with 17 correct answers, 1 unanswered question, and 7 incorrect, and also with 12 correct answers and 13 unanswered questions. There are three scores that can be obtained in exactly three ways. What is their sum?
Solve the following question in integers $$x^6 + 3x^3 +1 = y^4$$
Solve in positive integers the equation $x^2y + y^2z +z^2x = 3xyz$
Let $A$ and $B$ be two positive integers and $A=B^2$. If $A$ satisfies the following conditions, find the value of $B$:
- $A$'s thousands digit is $4$
- $A$'s tens digit is $9$
- The sum of all $A$'s digits is $19$
Is it possible to find four positive integers such that $2002$ plus the product of any two of them is always a square? If yes, find such four positive integers. If no, explain.
If the middle term of three consecutive integers is a perfect square, then the product of these three numbers is called a $\textit{beautiful}$ number. What is the greatest common divisor of all the $\textit{beautiful}$ numbers?
Find the smallest square whose last three digits are the same but not equal $0$.
Let $\overline{ABCA}$ be a four-digit number. If $\overline{AB}$ is a prime, $\overline{BC}$ is a square, and $\overline{CA}$ is the product of a prime and a greater-than-one square. Find all such $\overline{ABCA}$.
Let $A$ be a two-digit number, multiplying $A$ by 6 yields a three-digit number $B$. The difference of the two five-digit numbers obtained by appending $A$ to the left and right of $B$, respectively, is a perfect square. Find the sum of all such possible $A$s.
Find such a positive integer $n$ such that both $(n-100)$ and $(n-63)$ are square numbers.
Find such a positive integer $n$ such that both $(n+23)$ and $(n-30)$ are square numbers.
Find the smallest positive integer $n$ such that $\frac{12!}{n}$ is a square.
Consider the following $32$ numbers: $1!, 2!, 3!, \cdots, 32!$. If one of them is removed, then the product of the remaining $31$ numbers is a perfect squre. What is that removed number?