How would someone find a proof for this?

Question

Question:

Let $m \in Z$. Prove that if $m$ is the product of four consecutive integers, then $m + 1$ is a perfect square (that is, $m + 1 = k^{2}$ for some $k \in Z$).

Answer:
Since $m$ is a product of four consecutive integers, we may assume that $m=(k-1)k(k+1)(k+2)$ for some integer $k$. Thus,
$$m+1 = [k(k+1)][(k-1)(k+2)]+1=(k^{2}+k)(k^{2}+k-2)+1=(k^{2}+k)^{2}-2(k^{2}+k)+1=(k^{2}+k-1)^{2}$$

My question is, how would someone find a proof like this, why starting with $k-1$ and not $k$ or $k+2$, is it just bruteforce? and if we make the first choice "right", how would one deduce that we can rewrite the product as $(k^{2}+k-1)^{2}$, is it just a trick I should know? I mean I probably could solve this exercise, but it would take me a whole day of trial and error, maybe more. Would this me normal?

Answer 1

1. You can start with $k$ or $k+2$ or $k+100$ instead of $k-1$. Just the formulas will be a little longer.

2. You need to recognize the square of a sum $(a+b)^2=a^2+2ab+b^2=a(a+2b)+b^2$ and expect that somewhere in your computation you should be using that.

Answer 2

My approach is somewhat different.

Of course, I am meta-cheating. That is, I am assuming that the following approach will work, else the problem would not have been posed.

$\displaystyle [(k)(k-1)(k-2)(k+1)] + 1$

may be re-written as

$\displaystyle [(k^2 - k) \times (k^2 - k - 2)] + 1.$

At this point, it seems reasonable to attempt to make things easy by letting $r = (k^2 - k)$.

This means that the original expression may be re-written as

$[(r) \times (r - 2)] + 1 = r^2 - 2r + 1 = (r - 1)^2.$

Answer 3

I would start with examples to find a pattern
$1\cdot2\cdot3\cdot4+1=25 =5\cdot5$
$2\cdot3\cdot4\cdot5+1=121=11\cdot11$
$3\cdot4\cdot5\cdot6+1=361=19\cdot19$
$4\cdot5\cdot6\cdot7+1=841=29\cdot29$
$5\cdot6\cdot7\cdot8+1=1681=41\cdot41$

So the association is
$1 -> 5$
$2 -> 5+6$
$3 -> 5+6+8$
$4 -> 5+6+8+10$
$5 -> 5+6+8+10+12$
$n -> 5+6+8+10+12+...+(2n+2)$

and we have $5+6+8+10+12+...+(2n+2)=-1+2+4+6+8+10+12+...+(2n+2)=-1+(n+1)(n+2)$
Hence
$n(n+1)(n+2)(n+3)+1=[(n+1)(n+2)-1]^2$
or substituting n by n-1
$(n-1)n(n+1)(n+2)+1=(n^2+n-1)^2$.
After that it is easy to adjust to your prove.

Answer 4

Take $a = n + \frac{1}{2} $ for your favorite integer $n$. We get four consecutive integers as $$a - \frac{3}{2} \; , \; \; a - \frac{1}{2} \; , \; \; a + \frac{1}{2} \; , \; \; a + \frac{3}{2} $$ The product is $$ \left(a^2 - \frac{9}{4} \right) \left(a^2 - \frac{1}{4} \right) = a^4 - \frac{5}{2}a^2 +\frac{9}{16} $$ Adding $1$ to the product gives $$ a^4 - \frac{5}{2}a^2 +\frac{25}{16} = \left(a^2 - \frac{5}{4} \right)^2 $$

Finally, as $a=n +(1/2) ,$ we see $a^2 = n^2 + n + \frac{1}{4}, $ so that $$a^2 - \frac{5}{4} = n^2 + n - 1 $$ is an integer, and $1$ plus the product of four is $ \left(n^2 + n - 1\right)^2 $