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.

Question

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 * 4 * 5 * 3 = 120

Which one less than 121 a perfect square.
So how would I justify it I did
Let n be a integer

n (n + 1) (n + 2) (n + 3) + 1 = (m)^{2}

But I am not sure how to proceed.

Answer 1

Explanation:

\begin{aligned} n (n + 1) (n + 2) (n + 3) & = n^{4} + 6 n^{3} + 11 n^{2} + 6 n + 1 - 1 \\ = n^{2} (n^{2} + 6 n + 11 + \frac{6}{n} + \frac{1}{n^{2}}) - 1 \\ = n^{2} ({(n + \frac{1}{n})}^{2} + 6 (n + \frac{1}{n}) + 11 - 2) - 1 \\ = n^{2} ({(n + \frac{1}{n})}^{2} + 6 (n + \frac{1}{n}) + 9) - 1 \\ = n^{2} {(n + \frac{1}{n} + 3)}^{2} - 1 \\ = {(n^{2} + 3 n + 1)}^{2} - 1 \end{aligned}

Answer 2

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 (n + 1) (n + 2) (n + 3)

.

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.

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