Ackbach said:Couple of hints.
1. $n(n^2-1)(n+2)=(n-1)\, n \,(n+1)(n+2)$.
2. By the Quotient Remainder Theorem, there exist integers $q$ and $r$ such that $n=4q+r$, where $0\le r<4$.
Does this suggest anything to you?
tmt said:Right, so $(n-1)\, n \,(n+1)(n+2)$ is 4 consecutive integers. I get that if you take any arbitrary integer, if it is not divisible by 4, you can increment it by some int $c$ such that $0< c < 4$ and get an int that is divisible by 4.
Therefore, if you have 4 consecutive integers, one of those integers will be divisible by 4, and as a result the product of those 4 integers will be divisible by 4. I just don't know how to state or prove this formally.
The Quotient Remainder Theorem is a mathematical principle that states that when dividing one integer by another, the result can be expressed as a quotient and a remainder. For example, when dividing 13 by 5, the quotient is 2 and the remainder is 3.
The Quotient Remainder Theorem is often used in problem-solving to simplify large division problems. Instead of having to perform a long division, the theorem allows us to quickly determine the quotient and remainder without the need for extra calculations.
This theorem can only be used for division of integers, meaning whole numbers that can be positive, negative, or zero. It cannot be applied to division involving decimals or fractions.
A quotient is the result of division, while a remainder is the amount left over after dividing. In other words, the quotient tells us how many times the divisor can fit into the dividend evenly, while the remainder is the leftover amount that cannot be divided evenly.
Yes, the Quotient Remainder Theorem can be applied to any number of integers. For example, if we want to divide 50 by 6, the quotient is 8 with a remainder of 2. We can then take that remainder of 2 and continue dividing it by another integer to get another quotient and remainder.