# Proving the product of four consecutive integers, plus one, is a square

I need some help with a Proof:

Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.

I tried a direct proof where I said:

Assume $m$ is the product of four consecutive integers.
If $m$ is the product of four consecutive integers, then write $m=x(x+1)(x+2)(x+3)$ where $x$ is an integer.
Then $m=x(x+1)(x+2)(x+3)=x^4+6x^3+11x^2 +6x$.
Adding $1$ to both sides gives us:
$m+1=x^4+6x^3+11x^2+6x+1$.

I'm unsure how to proceed. I know I'm supposed to show $m$ is a perfect square, so I should somehow show that $m+1=a^2$ for some $a\in\mathbb{Z}$, but at this point, I can't alter the right hand side of the equation to get anything viable.

• Try something like $a=x^2+rx+s$ for suitable $r$, $s$. – Angina Seng Jun 26 '18 at 21:17
• Look at the first few square roots. There must be a simple quadratic that gives them. – lulu Jun 26 '18 at 21:19
• Well, this is a spoiler. $x^4 + 6x^3 + 11x^2 + 6x + 1 = (x^2 + 3x + 1)^2$ – fleablood Jun 26 '18 at 23:10
• Hint: $(x^2 + bx + c)^2 = x^4 + 2bx^3 + (2c + b^2)x^2 + 2bcx + c^2$. If $c^2 = 1; 2bc = 6; (2c+b^2) = 11; 2b=6$ has integer solution, you are golden. – fleablood Jun 26 '18 at 23:25
• – Ben Millwood Jun 27 '18 at 14:18

By the way, you mean $m+1 = x^4 + 6x^3 + 11x^2 + 6x + 1$.

Let's break it down. Obviously, you need a quadratic that, when squared, gives the above. How do you construct this? You need 3 numbers $a, b, c$ for $ax^2 + bx + c$.

Looking at the 4th degree thing above, what can you tell me about $a$, right off the bat? What about $c$? Then can you use this to tell you what $b$ is?

(I feel like I should not go further because I want you to solve it, but let me know if you need a clarification)

• Your result is right, of course, but you silently took a shortcut: nothing says a priori that the quartic polynomial is the square of a quadratic polynomial in the same variable. About all you have at this stage is that it is an integer function of $n$, asymptotic to $n^2$. – Yves Daoust Jun 28 '18 at 13:49
• This is a good point. I justified it by noticing the symmetry in the coefficients, so this can elucidate the approach. – BRSTCohomology Jun 28 '18 at 13:50

To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.

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

OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?

Let's throw in a simple example: $1\times2\times3\times4$. Notice how $1\times4=4$ and $2\times3=6$.

What about $2\times3\times4\times5$? This time $2\times5=10$ and $3\times4=12$.

It looks like the product of the "outer" pair is $n-1$ and the product of the "inner pair" is $n+1$.

Now we know how to attack this.

Let $k$ be some integer and $m=k(k+1)(k+2)(k+3)$.

\begin{align} m &= k(k+1)(k+2)(k+3) \\ &= (k+1)(k+2)\times(k(k+3)) \ \text{ (collecting inner and outer terms)}\\ &= (k^2 + 3k + 2) \times (k^2 + 3k) \\ &= ((k^2 + 3k + 1) + 1) ((k^2 + 3k + 1) - 1) \\ &= (k^2 + 3k + 1) ^ 2 - 1 \qquad \text{ (since }(a+b)(a-b)=a^2-b^2\text{)}. \end{align}

Since $k$ is an integer, $k^2 + 3k + 1$ is an integer so $m+1$ is a perfect square.

Write the product of the four consecutive integers starting at some $n-1$, so that $$m=(n-1)n(n+1)(n+2)+1,$$ and expand: \begin{align} m&=(n^2-1)(n^2+2n)+1=n^2(n^2-1)+2n(n^2-1)+1 \\ &=(n^2-1)^2+2n(n^2-1)+\not 1+ n^2{-}\!\not1 \\ &= \bigl((n^2-1)+n\bigr)^2. \end{align}

• The maniac downvoter struck again! As usual, without any explanation… – Bernard Jun 27 '18 at 8:13
• This is exactly how I solved it. Upvoting. – Cuspy Code Jun 27 '18 at 9:40
• Bizarre downvote; nice approach. Might be worth comparing to Zeitz's... +1 – Benjamin Dickman Jun 27 '18 at 23:30

See Part II there for several approaches taken by students (most of which are covered by other answers here; but the presentation is somewhat different).

One of the methods mentioned there is observing the symmetry above around $x= -3/2 = -1.5$, but then using this to inform a substitution: let $z = x + 1.5$ so that we have:

$$x(x+1)(x+2)(x+3) = (z-1.5)(z-0.5)(z+0.5)(z+1.5) = (z^2 - 1.5^2)(z^2 - 0.5^2)$$

Noting that $1.5^2 = 2.25$ and $0.5^2 = 0.25$, we could use one more substitution of $w = z^2 - 2.25$ to rewrite the final expression above as $w(w+2) = w^2 + 2w$, from which the addition of $1$ yields $(w+1)^2$ as desired. One can now rewrite in terms of just $x$ to finish matters off.

I think that the idea of the symmetry here is an important takeaway; incidentally, the problem is also broached in an exploratory manner at the beginning of Paul Zeitz's (2006) The Art and Craft of Problem Solving as Example 1.2.1.

• This is my preferred solution... because that's how I just solved it... However, I'm not a fan of using decimal approximations to represent fractions. Particularly for a number theory style problem. In my experience, it leads to too much confusion among students. It requires students to remember one more thing (this is an exact decimal representation) which is too much of an additional cognitive load. – John Jun 27 '18 at 22:37
• @John Definitely a worthy consideration: I put some related comments about this (specifically about notation) at MESE. But, I think you're right that fractions could make this answer cleaner. . . – Benjamin Dickman Jun 27 '18 at 23:26
• @John You can exploit the symmetry without using decimals, just keep fractions, multiple by suitable integer ($16$ in this case) and you are back with integer coefficients. You will get $16f\left(x+\frac{1}{2}\right)=(4x^2-5)^2$, which in turn yields $16f(x)=(4x^2-4x-4)^2$, and you are basically done. – Sil Jun 28 '18 at 7:59

Given $m$ is the product of four consecutive integers. $$m=p(p+1)(p+2)(p+3)$$where $p$ is an integer

we need to show that $p(p+1)(p+2)(p+3)+1$ is a perfect square

Now,$$p(p+1)(p+2)(p+3)+1=p(p+3)(p+1)(p+2)+1$$ $$=(p^2+3p)(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+2)-(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+1+1)-(p^2+3p+2)+1$$ $$=(p^2+3p+1)(p^2+3p+1)+(p^2+3p+1)-p^2-3p-2+1$$ $$=(p^2+3p+1)(p^2+3p+1)=(p^2+3p+1)^2$$ So, $m+1$ is a perfect square where $m$ is the product of four consecutive integers.

Proof without words:

$\hspace{2cm}$ $$\color{red}x\color{blue}{(x+1)(x+2)}\color{red}{(x+3)}+1=\color{red}{(x^2+3x)}\color{blue}{(x^2+3x+2)}+1=(x^2+3x+1)^2.$$

Note that $f(x)=x(x+1)(x+2)(x+3)+1$ is a degree $4$ polynomial with leading term $x^4$ and symmetric around $x=-\frac32$. We might try the ansatz $f(x)=g(x)^2$ with $g(x)=x^2+px+q$ because then the leading term of $g(x)^2$ is also $x^4$. We suspect that $g$ is also symmetric around $x=-\frac32$ and hence write it as $g(x)=(x+\frac32)^2+c$. Note that $f(0)=f(-1)=1$, so we want $g(0)=\pm1$ and $g(-1)=\pm1$. The first means $c\in\{-\frac54,-\frac{13}4\}$, the second means $c\in\{\frac34,-\frac54\}$. We conclude that $c=-\frac54$. Without further calculation, we see that $f(x)-g(x)^2=0$ for $x=0$ and $x=-1$ and by symmetry also for $x=-2$ and $x=-3$. As the leading terms cancel, the polynomial $f(x)-g(x)^2$ is in fact of degree of at most $3$. Having four distinct roots, it must be identically zero, as desired.

\begin{eqnarray*} m=n(n+1)(n+2)(n+3) =n^4+6n^3+11n^2+6n. \end{eqnarray*} So \begin{eqnarray*} m+ 1 =n^4+6n^3+11n^2+6n+1=(n^2+3n+1)^2. \end{eqnarray*}

• Judging by the way in which the problem is asked: It might be helpful to the OP to explain how to get that factorization!  For example, noting that $n(n+3) = n^2 + 3n := m$ and $(n+1)(n+2) = n^2 + 3n + 2 = m+2$, so that the product plus one is $m^2 + 2m + 1 = (m+1)^2 = (n^2 + 3n + 1)^2$ as you observed. – Benjamin Dickman Jun 26 '18 at 22:17
• What is the thought process used here? – richard1941 Jul 7 '18 at 2:25

The polynomial $x^4+6x^3+11x^2+6x+1$ has symmetric coefficents - more precisely, it's called a palindromic polynomial:

$$p(x)=x^4+ax^3+bx^2+ax+1$$

The goal is to factor $p(x)$ (and show that it factors to a square of some expression). Let's start by dividing by $x^2$ and refactoring:

$$\frac{p(x)}{x^2} = q(x) = x^2+\frac{1}{x^2}+a\left(x+\frac{1}{x}\right)+b$$

It's tempting to make a substitution $y=x+\frac{1}{x}$:

$$q(x) \rightarrow q(y) = y^2+ay+(b-2)$$

This: $q(y)=0$, being a quadratic equation, is something that can be automatically solved:

$$y_{1,2}=\frac{-a\pm \sqrt{a^2-4(b-2)}}{2}$$

and inserting the values $a=6$ and $b=11$, one obtains (note the expression under the square root is equal to zero):

$$y_1=y_2=-3$$

so that $q(y)=(y+3)^2$ - at this point one sees it's a perfect square, and due to $q(y)=\frac{p(x)}{x^2}$, basically we're done with the proof at this point.

Just to take things to their end:

Because $q(y)=0 \Leftrightarrow q(x)=0 \Leftrightarrow p(x)=0$:

$$q(y) = (y+3)^2 = \left(x+\frac{1}{x}+3\right)^2 = \frac{(x^2+3x+1)^2}{x^2} = \frac{p(x)}{x^2}$$

hence $x^4+6x^3+11x^2+6x+1 = (x^2+3x+1)^2$ - a perfect square indeed.

Empirically:

Consider the function

$$p(n):=\sqrt{n(n+1)(n+2)(n+3)+1}.$$

For $n=0,1,2,3,\cdots$ we have $p(n)=1,5,11,19,29,\cdots$ a sequence with constant second order differences ($2$), and we can postulate the polynomial

$$n^2+3n+1$$

(because $p(0)=1$, the coefficient of $n^2$ must be $1$, and $p(n)-n^2-1=0,3,6,9,\cdots$)

Now the identity

$$n(n+3)(n+1)(n+2)=(n^2+3n)(n^2+3n+2) \\=(n^2+3n+1-1)(n^2+3n+1+1) \\=(n^2+3n+1)^2-1.$$

becomes apparent.

Another approach is by bringing more symmetry and shifting the variable by $3/2$.

$$\sqrt{\left(m-\frac32\right)\left(m-\frac12\right)\left(m+\frac12\right)\left(m+\frac32\right)+1} =\sqrt{\left(m^2-\frac94\right)\left(m^2-\frac14\right)+1} =\sqrt{m^4-\frac52m^2+\frac{25}{16}}=m^2-\frac54,$$

which is

$$n^2+2\frac32n+\left(\frac32\right)^2-\frac54.$$

If the numbers are $x, x+1, x+2, x+3$.

Let $\frac m2 = x+1.5$ be the midpoint of the four consecutive integers, so that the integers are $\frac {m-3}2, \frac {m-1}2, \frac {m+1}2, \frac {m+3}2$. (Note: $m$ is odd and $\frac m2$ is not an integer.)

So $x(x+1)(x+2)(x+3) + 1 =$

$\frac {(m-3)(m+3)(m-1)(m+1)}{16} + 1=$

$\frac {(m^2 -9)(m^2 - 1) + 16}{16} =$

$\frac {(m^2-10m^2 + 9) +16}{16} =\frac {m^2-10m^2 + 25}{16}=$

$(\frac {m^2 -5}{4})^2$.

Now $m$ is odd. So let $m = 2n+1$ then

$x(x+1)(x+2)(x+3) + 1 = (\frac {(2n+1)^2 -5}{4})^2=$

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

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

====

So ....

If $x$ is the first integer and $\frac {2n + 1}2 = x + \frac 32$ then

$n = x + 1$.

So $x(x+1)(x+2)(x+3) + 1 = ((x+1)^2 + (x+1) - 1)^2 = (x^2 + 3x +1)^2$

....

So as you got $x^4 + 6x^3 + 11x^2 + 6x + 1$ that actually equals $(x^2 + 3x +1)^2$

Indeed $x^4 + 6x^3 + 11x^2 + 6x + 1 = x^2(x^2 + 3x + 1) + 3x^3 +10x^2 + 6x + 1$

$= x^2(x^2 + 3x+ 1) + 3x(x^2 + 3x + 1) +x^2 +3x + 1$

$= (x^2 + 3x + 1)^2$.

D'oh.

If $x(x+1)(x+2)(x+3) + 1 = a^2$ then

$x(x+ 1)(x+ 2)(x+3) = a^2 - 1 = (a + 1)(a-1)$

To get factors that close together they'd have to be

$a = x(x+3) \pm 1= (x+ 1)(x+2) \mp 1$

and indeed $a= x(x+3) + 1 = (x+1)(x+2) - 1= x^2 +3x + 1$.

proves the statement! (If we work backwards.)

Suppose that you have 4 consecutive numbers $a, b, c, d$. They can be expressed as $a=t-\frac{3}{2}$, $b=t-\frac{1}{2}$, $c=t+\frac{1}{2}$ and $d=t+\frac{3}{2}$ for some number $t$.

Now, $$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = t^2 - \left(\frac{3}{2}\right)^2 = t^2 - \frac{9}{4}$$ and $$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = t^2 - \left(\frac{1}{2}\right)^2 = t^2 - \frac{1}{4}$$

If we define $y = t^2 - \frac{5}{4}$, we have $$ad = \left(t-\frac{3}{2}\right)\left(t+\frac{3}{2}\right) = y - 1$$ and $$bc = \left(t-\frac{1}{2}\right)\left(t+\frac{1}{2}\right) = y + 1$$

Furthermore, since the LHS in both cases is an integer, it is clear that $y$ is an integer.

So the product of all four numbers is $$abcd = (y - 1)(y+1) = y^2 - 1$$ one less than the square of an integer.

This is another way of looking at @jwg's answer. Let the four consecutive numbers be $a,b,c,d$ and let $t$ be the number half-way between $b$ and $c$. Then

\begin{align} abcd + 1 &= \bigg(t-\frac 32\bigg)\bigg(t-\frac 12\bigg) \bigg(t+\frac 12\bigg)\bigg(t+\frac 32\bigg) + 1\\ &= \bigg(t^2-\frac 94\bigg)\bigg(t^2-\frac 14\bigg) + 1\\ &= t^4 - \frac 52 t + \frac{25}{16}\\ &= \bigg(t^2 - \frac 54 \bigg)^2 \end{align}

Letting $t = a + \frac 32$, we get

\begin{align} abcd + 1 &= \bigg(t^2 - \frac 54 \bigg)^2 \\ &= \bigg(a^2 +3a + \dfrac 94 - \frac 54 \bigg)^2 \\ &= (a^2 +3a + 1)^2 \end{align}

This suggests the following solution.

\begin{align} abcd + 1 &= a(a+1)(a+2)(a+3) + 1 \\ &= a(a+3) \cdot (a+1)(a+2) + 1 \\ &= (a^2+3a) (a^2+3a+2) + 1 \\ &= (a^2+3a+1 \ - \ 1)(a^2+3a+1 \ + \ 1) + 1 \\ &= (a^2+3a+1)^2 - 1 + 1 \\ &= (a^2+3a+1)^2 \end{align}