# Discrete math number problem. How would I justify the following statement. Two integers are consecutive if and only if one is more than the other. Any product of four consecutive integers is one less than a perfect square.

Discrete math number problem
How would I justify the following statement.
Two integers are consecutive if and only if one is more than the other. Any product of four consecutive integers is one less than a perfect square.
I think this is true.
because for example
$2<3<4<5$
$2\ast 4\ast 5\ast 3=120$
Which one less than 121 a perfect square.
So how would I justify it I did
Let n be a integer
$n\left(n+1\right)\left(n+2\right)\left(n+3\right)+1=\left(m{\right)}^{2}$
But I am not sure how to proceed.
You can still ask an expert for help

## Want to know more about Discrete math?

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

juicilysv
Explanation:
$\begin{array}{rl}n\left(n+1\right)\left(n+2\right)\left(n+3\right)& ={n}^{4}+6{n}^{3}+11{n}^{2}+6n+1-1\\ & ={n}^{2}\left({n}^{2}+6n+11+\frac{6}{n}+\frac{1}{{n}^{2}}\right)-1\\ & ={n}^{2}\left({\left(n+\frac{1}{n}\right)}^{2}+6\left(n+\frac{1}{n}\right)+11-2\right)-1\\ & ={n}^{2}\left({\left(n+\frac{1}{n}\right)}^{2}+6\left(n+\frac{1}{n}\right)+9\right)-1\\ & ={n}^{2}{\left(n+\frac{1}{n}+3\right)}^{2}-1\\ & ={\left({n}^{2}+3n+1\right)}^{2}-1\end{array}$
###### Not exactly what you’re looking for?
Macioccujx
Explanation:
Try to proceed heuristically: write down the numbers (possible squares) you get for the first few values of n. Write down the numbers they are the squares of. Try to find a relationship between this and n. Is the former a polynomial in n? Develop the product $n\left(n+1\right)\left(n+2\right)\left(n+3\right)$.