What do you think about it?

[vict85]

It’s a fact or a conjecture? I mean, I have no idea it's correct or not. In the case it is a conjecture, did you test it for at least 1000 cases?

[vict85]

OK, it is true. In fact \(\displaystyle n(n+1)(n+2)(n+3) + 1 = (n(n+3) + 1)^2 \).

Proof:
\begin{align} n(n+1)(n+2)(n+3) + 1 &= n^2(n+3)(n+1) + 2n(n+3)(n+1) + 1 \\
&= n^2(n+3)(n+1) + 2n(n+3)(n+1) + 1 \\
&= n^2(n+3)(n+1) + 2n^2(n+3) + 2n(n+3) + 1 \\
&= n^2(n+3)^2 + 2n(n+3) + 1 \\
&= (n(n+3) + 1)^2 \\
\end{align}

[Pianoth]

"A brain box" means "an intelligent person", so you should use some verb Anyway, that's a nice exercise from an entrance examination, as far as I know (I searched on the internet )

[Pianoth]

Don't worry, I didn't say anything bad about your english
Let me say something related:
The product of 4 consecutive odd numbers plus 16 is always a perfect square.
Some examples:
$11*13*15*17 + 16 = 36481 = 191^2$
$3*5*7*9 + 16 = 961 = 31^2$
$15*17*19*21+16 = 101761 = 319^2$
What do you think of this? It's a very similar result, but I haven't found it on the internet.

[vict85]

You only have to factorize the polynomial \(\displaystyle f(k) = (2k+1)(2k+3)(2k+5)(2k+7) + 16 \)...

[vict85]

So, actually, they doesn't need to be odd...

[Pianoth]

I know, but the result seems more impressive

[Pianoth]

I explored $x(x+a)(x+2a)(x+3a)$ and look at that:
\(x(x+1)(x+2)(x+3)+1=(1 + 3 x + x^2)^2\)
\(x(x+2)(x+4)(x+6)+16=(4 + 6 x + x^2)^2\)
\(x(x+3)(x+6)(x+9)+81=(9 + 9 x + x^2)^2\)
\(x(x+4)(x+8)(x+12)+256=(16+12 x+x^2)^2\)
\(\vdots\)
\(x(x+a)(x+2a)(x+3a)+a^4=(a^2 + 3 a x + x^2)^2\)

Initially I was even surprised, but actually this is quite obvious:
$x(x+a)(x+2a)(x+3a)=6 a^3 x + 11 a^2 x^2 + 6 a x^3 + x^4$
Some term is visibly missing, you see $x, x^2, x^3, x^4, a, a^2, a^3$ but you don't see $a^4$.