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# 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
3 years ago

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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 $$(x-1)^2$$. · 2 years, 11 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. · 2 years, 11 months ago

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 $$abcd+1$$. · 2 years, 11 months ago