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Relevant wiki: Algebraic Manipulation - Substitution

Another approach: $\begin{aligned} n(n+1)(n+2)(n+3) + 1 &= n(n+3)(n+1)(n+2) + 1 \\ &= (n^2+3n)(n^2+3n+2) + 1 \qquad \small \text{[Now, letting } N=n^2+3n+1] \\ &= (N-1)(N+1) + 1 \\ &= N^2 \end{aligned}$ and we have the desired result.

Relevant wiki: Algebraic Manipulation - Rearranging

Let $p$ the integer and we have that $N = 1+{\color{#3D99F6} p}(p+1)(p+2){\color{#3D99F6}(p+3)} \\ N = 1 +{\color{#3D99F6}(p^2+3p)}(p^2+3p+2) \\ N= 1+ ({\color{#3D99F6}p^2+3p+1}-1)({\color{#3D99F6}p^2+3p+1} +1) = (p^2+3p+1)^2$ Hence, number obtained is always perfect square number.

$\begin{aligned} 1+(p-1)(p)(p+1)(p+2) &= 1 + (p^3 - p)(p+2) \\ &= p^4 + 2p^3 - p^2 - 2p + 1\\ &= (p^2 + p - 1)^2 \end{aligned}$

New solution

Add two lines to the top of the list: $\begin{aligned} 1 + (-1)\times 0 \times 1 \times 2 & = (-1)^2 \\ 1 + 0 \times 1 \times 2 \times 3 & = 1^2 \\ 1 + 1 \times 2 \times 3 \times 4 & = 5^2 \\ 1 + 2\times 3 \times 4 \times 5 & = 11^2 \\ 1 + 3\times 4 \times 5 \times 6 & = 19^2\end{aligned}$ The left-hand side may be written as $f(n) = 1 + n(n+1)(n+2)(n+3)$ with $n = -1, \dots, 4$ . This is a fourth-degree polynomial.

The right-hand side may be written as $g(n) = (h(n))^2$ . Note the second differences of $h(n)$ : $\begin{array}{ccccccccc} -1 && 1 && 5 && 11 && 19 \\ & 2 && 4 && 6 && 8 \\ && 2 && 2 && 2\end{array}$ Since the second differences are constant, $h(n)$ is a quadratic polynomial. This makes $g(n)$ a fourth-degree polynomial.

Finally, since $f$ and $g$ are fourth-order polynomials that agree in five points, they must be identical. Thus $f(n) = g(n)$ for any possible value of $n$ .

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

$= 1 + (n - 1)(n + 1)n(n + 2)$

$= 1 + (n^2 - 1)(n^2 + 2n)$

$= -n^2 + 1 + n^2 + (n^2 - 1)(n^2 + 2n)$

$= -(n^2 - 1) + n^2 + (n^2 - 1)(n^2 + 2n)$

$= n^2 + (n^2 - 1)(n^2 + 2n - 1)$

$= n^2 + (k - n)(k + n)$ where $k = n^2 + n - 1$

$= n^2 + k^2 - n^2$

$= k^2$

Note that the general expression is of the form:

$P(n) = 1+n(n+1)(n+2)(n+3)(n+4) = n^4+6n^3+11n^2+6n+1$

Where $n = 1, 2, 3$ are provided. We are looking for an equation of the form: $Q(n, B) = (n^2+Bn+1)^2$ such that the first and last coefficients line up between P and Q: $(n^2)^2 = n^4$ and $1^2 = 1$

$Q(n, B) = (n^2+Bn+1)^2 = n^4+2Bn^3+(B^2+2)n^2+2Bn+1$

Note that: for there to exist a B such that this squared polynomial exists for the given polynomial, it must be the case that that, for a given B, $P(n) - Q(n, B) = 0$

$(n^4+6n^3+11n^2+6n+1) - (n^4+2Bn^3+(B^2+2)n^2+2Bn+1) = (6-2B)n^3+(9-B^2)n^2+(6-2B)n = 0$

For this to be true for an arbitrary n, we must find a B such that $6-2B =0$ , $9-B^2 =0$ , and $6-2B = 0$ are all simultaneously satisfied for our respective $n^3$ , $n^2$ , and $n$ terms.

Fortunately, solving for any of these show that $B = 3$ is the only solution.

$P(n) = Q(n, 3) = (n^2+3n+1)^2$

Because the integers form a commutative ring, $n\epsilon \mathbb{Z} \rightarrow \sqrt{P(n)}\epsilon \mathbb{Z}$

Thus, the sum of 1 and four arbitrary consecutive integers will always result in another integer!

Another way of looking at this is recognizing that $n(n+3) = (n+1)(n+2) - 2$ . So this can be regrouped as $(n+1)(n+2)[(n+1)(n+2) - 2] + 1$ . For simplification, let $(n+1)(n+2) = x$ , then this becomes $x(x-2) +1 = x^2 -2x + 1$ , which is $(x-1)^2$ . Therefore, this is always a square.

I know no math, so I present you the approach of my intuition, which is not a formal proof by any means, but it is something to think about, and maybe something that can help someone to intuitively understand something more deep about the numbers, than can be missed by looking only at formalisms.

Look close at the 4 numbers we are multiplying. Firstly multiply the first and the last one only, write down the result, then multiply the second and the third one, and write down the result. You will get:

$1+1*2*3*4=5^2$ | Results: 4 and 6

$1+2*3*4*5=11^2$ | Results: 10 and 12

$1+3*4*5*6=19^2$ | Results: 18 and 20

Do you see the pattern? All produced results will be apart by 2, this is the first interesting thing to think about, and after some thought about the process of multiplication, you will intuitively understand why is that the case. Now, what do you get when you will multiply two numbers with a difference of 2? You will always get a number that is only 1 short of a perfect square. In fact, by going: $1*3+1$ , $2*4+1$ ... you will get the ALL perfect squares. This is also something interesting to think about, because it represents a more basic feature of the numbers.

These facts only made me sure of the correct solution, even though I have no knowledge of how to describe the solution formally. This is just something to think about, going beyond formalisms. I hope it will be interesting for someone at least.

Start with the four products n(n + 1)(n + 2)(n + 3) leaving out the +1 for now

Expand the two middle terms: n(n^2 + 3n + 2)(n + 3) = n(n + 3)[n(n + 3) + 2]

Multiply the bracketed expression by n(n + 3):

n^2(n + 3)^2 + 2n(n + 3) and now add back the 1 that was left out before:

n^2(n + 3)^2 + 2n(n + 3) + 1 = [n(n + 3) + 1]^2 a perfect square.

Notice that $n\cdot (n+3) = n^2 + 3n$

and

$(n+1)\cdot (n+2) = n^2 + 3n + 2$ ,

so taking

$x= n^2+3n+1$ ,

we have that $x^2 -1 = (x+1)\cdot (x-1) = (n)\cdot (n+1)\cdot(n+2)\cdot (n+2)$ .

n(n+1)(n+2)(n+3) +1 = (n^2 + 3n + 1)^2. Ed Gray

The expression is $(y+1)(y+2)(y+3)(y+4)+1$

Using some simple algebra, this expression reduces to -

$y^4+10y^3+35y^3+50y+25$ which is equal to $(y^2+5y+5)^2$

Since $y^2+5y+5$ is an integer, we can say that the expression $(y+1)(y+2)(y+3)(y+4)+1$ is always a perfect square.

We can name the consecutive numbers as: $m, n, p\ and\ q.$ such that $m<n<p<q$ .

Let $k = \frac{ n+p}{2} = \frac{m+q}{2} = n+1 /2$

Then $m = k- 3/2,\ n = k- 1/2,\ p = k + 1/2,\ q = k + 3/2$

$N$ , which is the product plus one, is given by:

$N = (k-\frac 3 2)(k+\frac 3 2)(k-\frac 1 2)(k+\frac 1 2) +1$

$= (k^2-\frac 9 4)(k^2 - \frac 1 4) + 1\ =\ k^4 - \frac 5 2 k^2 + \frac 9 {16} + 1$

$= k^4 - \frac 5 2 + \frac {25}{16}\ =\ (k^2 - \frac 54)^2$

Subtituting back $k = n + \frac 12$ :

$N = ((n+\frac 12)^2 - \frac 54)^2$

$\ \ = \ (n^2 +n + \frac 14 - \frac 54)^2 \ =\ (n^2 +n -1) ^2$

SInce I don't enjoy expanding terms too much, I shifted everything by $\frac{1}{2}$ to get nice $(x + c)(x - c)$ expressions:

Let $a = k + \frac{1}{2}$ , with integer k. Then the product of four consecutive integers could be expressed as $(a - \frac{3}{2}$ ) $(a - \frac{1}{2}$ ) $(a + \frac{1}{2}$ ) $(a + \frac{3}{2}$ ) for some a. So:

$P + 1 = (a + \frac{1}{2})(a - \frac{1}{2})(a + \frac{3}{2})(a - \frac{3}{2}) + 1 = (a^2 - \frac{1}{4})(a^2 - \frac{9}{4}) + 1 = a^4 - \frac{5}{2}a^2 + \frac{9}{16} + 1 = a^4 - \frac{5}{2}a^2 + \frac{25}{16} = (a^2 - \frac{5}{4})^2$

Now we must show that $a^2 - \frac{5}{4}$ is an integer, and for that we can substitute in $a = k + \frac{1}{2}$ :

$(a^2 - \frac{5}{4})^2 = (k^2 + k + \frac{1}{4} - \frac{5}{4})^2 = (k^2 + k - 1)^2$

$k^2 + k - 1$ is clearly an integer, so P + 1 is a perfect square.

Examining the provided examples, we can notice that the root is the square of the second integer plus the first integer. A few more checks and we find that holds true for higher sequences, too:

$1 + 8 \times 9 \times 10 \times 11 = 89^2 = (9^2 + 8)^2$

Assuming this holds true, it makes our job of calculation easier - we don't have to find a perfect square amidst the quartic equation, instead, we just have to expand and compare that they are equal. Since the second digit is squared in the solution, I will use $x$ to represent it, and $(x-1)$ for the first term. It also provides nicer terms for multiplication later.

$1 + (x-1)x(x+1)(x+2) \overset{?}{=} (x^2 + (x-1))^2$

Multiply convenient terms on the left, $(x-1)(x+1)$ and $x(x+2)$ , and expand the right:

$1 + (x^2-1)(x^2+2x) \overset{?}{=} (x^2 + x - 1)(x^2 + x - 1)$

Multiply out the terms:

$1 + x^4 - x^2 + 2x^3 - 2x \overset{?}{=} ((x^4 + x^3 - x^2) + (x^3 + x^2 - x) + (-x^2 - x + 1)$

Combine and order like terms on each side to standard polynomial form, demonstrating equality:

$x^4 + 2x^3 - x^2 - 2x + 1 = x^4 + 2x^3 - x^2 - 2x + 1$

Since we show that the sequence of four is equal to the square of $x^2 + (x-1)$ , where $x$ is the second term, the answer is yes .

Note that examining the pattern on the right makes the algebra easier; we can use simple expansion instead of trying to figure out the root algebraically.

I simply approximated a function out of this by switching the factors for example for the Term 3x4x5x6 to 6x5x4x3 and this equals 6x5x4x3x2x1/2x1 or X!/(X-4)!

we don't want to include the 0 because this will give us 1 wich is equal to 1^2 and because there are 4 factors we will also get a result for the negative numbers

now we simply write the function f(x)= sqrt[1+(X!/(X-4)!)] if the function is a positive integer for each X then it's true that the equation always equals a perfect square. And indeed we get true as an result (you could either try typing on a calculator or write a short program)

$1 + j(j + 1)(j + 2)(j + 3) = 1 + (j^3 + 3j^2 +2j)(j + 3) = j^4 + 6j^3 + 2j^2 + 9j^2 + 6j + 1 =$

$j^4 + 2(3j + 1)j^2 + (3j + 1)^2 = (j^2 + 3j + 1)^2$ .

1+(X) (X+1) (X+2)*(X+3)=X^4+6X^3+11X^2+6X+1=(X^2+3X+1)^2 so must always be a square

let $n$ be our integer, then we have:

$\begin{aligned} n(n+1)(n+2)(n+3) + 1 &= n^4+6n^3+11n^2+6n+1 \qquad \small \text{[Now, spreading the elements to form the binomial coefficient]} \\ &= (n^4+4n^3+6n^2+4n+1)+(2n^3+5n^2+2n) \\ &= (n+1)^4+2n(n^2+2n+1)+n^2 \\ &= (n+1)^4+2n(n+1)^2+n^2 \qquad \small \text{[Now, letting } (n+1)^2=a] \\ &= a^2+2na+n^2 \\ &= (a+n)^2 \\ &= [(n+1)^2+n]^2 \end{aligned}$

Thus, 1 plus the product of any set of four consecutive integers will always be a perfect square.