# Prove that the root is a whole number

Show that

$\sqrt{1+n(n+1)(n+2)(n+3)}$

is a whole number for all whole numbers $n$.

I can see that there are four consecutive numbers, meaning that the expression can be written as $\sqrt{1+24m}$ Also, it is easy to see that the expression is true for $n=1$ but I can't get the induction to work...

Notice that

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

Let $n^2+3n=t$

then we have $$1+n(n+1)(n+2)(n+3)=t^2+2t+1=(t+1)^2$$

so $$\sqrt{1+n(n+1)(n+2)(n+3)} =t+1=n^2+3n+1$$ which is clearly a whole number for $n \in Z$

• +1, but I think $$1+n(n+1)(n+2)(n+3)=(n2+3n+1)^2$$ is detailed enough – miracle173 Jan 15 '17 at 23:51
• @miracle173 just wanted to make sure he/she understood it! sometimes it is hard to see those pesky quadratics! :) – bigfocalchord Jan 15 '17 at 23:52
• Thanks, I had a hunch that factorization would work but I couldn't get the terms to factor the way they were supposed to. – Björn Lindqvist Jan 15 '17 at 23:56
• @BjörnLindqvist no worries , if you think my answer was sufficient you can "tick" it. :) – bigfocalchord Jan 15 '17 at 23:57

Another approach is to note that $(n+1)(n+2)=n(n+3)+2$, so if $M=n(n+3)+1$ then $M-1=n(n+3)$ and $M+1=(n+1)(n+2)$ so $M^2-1=(M-1)(M+1)=n(n+1)(n+2)(n+3)$.

Okay, I hate multiplying and factoring. So I note:

If $m = n+ 1.5$ then

$1 + n(n+1)(n+2)(n+3) = 1+ (m- \frac 32)(m- \frac 12)(m+\frac 12)(m + \frac 32) = 1 + (m^2 - \frac 94)(m^2 - \frac 14)=$

$1 + \frac 9{16} - \frac {10}4 m^2 + m^4 =$

$\frac {25}{16} - \frac {10}4 m^2 + m^4=$

$(m^2 - \frac 54)^2 =$

$((n + \frac 32)^2 - \frac54)^2 =$

$(n^2 + 3n + \frac 94 - \frac 54)^2=$

$(n^2 + 3n + 1)^2$.

So $\sqrt {1 + n(n+1)(n+2)(n+3)} = n^2 + 3n + 1$ which is a whole number.

=====

So... was I lucky that it worked out? Maybe, but I don't think so.

I knew that if it was a perfect square, and if I could get it into quadratic terms $v^2 + bv^2 + c$ for $v$ somehow in terms of $n$ then it would have to be a perfect square.

And I knew that if I took the middle of the the $n,n+1,n+2,n+3$ and replaced with $(m-w)(m-u)(m+u)(m+w)$ I'd get $(m^2 - w^2)(m^4 - u^2)$ which would be a quadratic terms for $v = m^2 = (n+w)^2$.

So if it was a perfect square, I knew I'd have to have success. If it wasn't a perfect square I knew I'd fail.