# Four Consecutive Integers Plus 1

Let $$a, b, c, d$$ be four consecutive integers. Prove that $$abcd + 1$$ is a perfect square. Furthermore, prove that the the square root of $$abcd + 1$$ is equal to the average of $$ad$$ and $$bc$$.

Solution

Let $abcd + 1$ be represented as $(n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1$.

Factoring $abcd + 1$ yields ${({n}^{2}+n-1)}^{2}$, proving that it is a perfect square.

The average of $ad$ and $bc$ is

$\frac{1}{2}\left[(n-1)(n+2) + n(n+1)\right] = \frac{1}{2}(2{n}^{2} +2n -2 )= {n}^{2}+n-1$

which is the square-root of $abcd+1$.

Check out my other notes at Proof, Disproof, and Derivation

Note by Steven Zheng
6 years, 11 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

Solution

Let $abcd + 1$ be represented as $(n-1)(n)(n+1)(n+2) +1 = {n}^{4}+2{n}^{3}-{n}^{2}-2n + 1$.

Factoring $abcd + 1$ yields ${({n}^{2}+n-1)}^{2}$, proving that it is a perfect square.

The average of $ad$ and $bc$ is

$\frac{1}{2}\left[(n-1)(n+2) + n(n+1)\right] = \frac{1}{2}(2{n}^{2} +2n -2 )= {n}^{2}+n-1$

which is the square-root of $abcd+1$.

- 6 years, 10 months ago

It seems like a lot of work to expand $(n-1)(n)(n+1)(n+2)+1$ and factor it. Here's a cool trick.

Rearrange it like this,

$((n-1)(n+2))((n)(n+1))+1\cdots (1)$

Now let $n^2+n=x$.

$(1)$ simplifies to $(x-2)(x)+1$ which is equal to $(x-1)^2$.

- 6 years, 9 months ago

Nice! I solved this a long, long, time ago and I copied this from my notebook. Back then, I was just learning algebra, and brute forced these things.

- 6 years, 9 months ago