0
$\begingroup$

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...

$\endgroup$
9
$\begingroup$

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 $n \in Z$

$\endgroup$
  • $\begingroup$ +1, but I think $$1+n(n+1)(n+2)(n+3)=(n2+3n+1)^2$$ is detailed enough $\endgroup$ – miracle173 Jan 15 '17 at 23:51
  • $\begingroup$ @miracle173 just wanted to make sure he/she understood it! sometimes it is hard to see those pesky quadratics! :) $\endgroup$ – bigfocalchord Jan 15 '17 at 23:52
  • $\begingroup$ Thanks, I had a hunch that factorization would work but I couldn't get the terms to factor the way they were supposed to. $\endgroup$ – Björn Lindqvist Jan 15 '17 at 23:56
  • $\begingroup$ @BjörnLindqvist no worries , if you think my answer was sufficient you can "tick" it. :) $\endgroup$ – bigfocalchord Jan 15 '17 at 23:57
4
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

I had success.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.