The Perfect Square Secret of Consecutive Integers // Math Minute [#75]

Consecutive integers have all kind of interesting properties and are therefore often the basis for algebraic equations and word problems. But those problems are usually about the sums or differences of consecutive integers. However, there's a particularly interesting property you can discover about four consecutive integers if you multiply them rather than add or subtract them.
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The product of any four consecutive integers plus one will ALWAYS be a perfect square. This property is the basis for a lot of math contest questions, especially if you're expected to simplify some sort of crazy square root without a calculator.
The way the property works is this. Consider four consecutive integers n, n+1, n+2, and n+3. The product of these four integers is n × (n+1) × (n+2) × (n+3). If we add one to that, we're looking at n(n+1)(n+2)(n+3) + 1.
To understand why this sum is always a perfect square, let's expand and simplify it:
Start by pairing the middle two terms and outer two terms of the product:
n(n+1)(n+2)(n+3) + 1 = n(n+3) × (n+1)(n+2) + 1
This can be rewritten as:
(n² + 3n) × (n² + 3n + 2) + 1
With either a further expansion or some clever substitution (as in the video), we can re-write that expression as the perfect square of n² + 3n + 1. This means that if you begin with an integer like 7, the product of 7, 8, 9, and 10 + 1 must be the same as 7² + 3(7) + 1. Since that value is 5041, and 5041 is 71², we can see that it works!
To see this property in action, check out these related videos:
+ Simplifying a Crazy Radical Part 1: • Simplifying a Radical ...
+ Simplifying a Crazy Radical Part 2: • Simplifying a Radical ...
+ Mental Math Challenge: • mental math: for when ...
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[@muddathiryousif2004]

That's actually smart and very interesting 😂❤
Amma save it

[@FranfranNotAMan]

As a math tutor I love this kinda stuff