/sci/ - Science & Math

[Anonymous No.10531972]

Checkmate, /sci/shits.

[Anonymous No.10531995]

you got me.

Now do your own homework faggot.

[Anonymous No.10532030]

Dude just use your ABCs just go on abcmouse nigga

[Anonymous No.10532043]

>>10532030
kek

[Anonymous No.10532075]

>>10531972
n(n+1)(n+2)(n+3) + 1 = (n^2 + 3n + 2)(n^2 + 3n) + 1
= (n^2 + 3n + 1)(n^2 + 3n) + (n^2 + 3n) + 1 = (n^2 + 3n + 1)(n^2 + 3n + 1)
= (n^2 + 3n + 1)^2

[Anonymous No.10532084]

>>10531972
My god is this what passes for pre-algebra in American schools these days.??

[Anonymous No.10532088]

>[math]n \in \mathbb R[/math]
REEEEEEEEEEEEEEEEEEEEEEEEEEE

[Anonymous No.10532090]

>>10532075
Is there a less crude way to do it?

[Anonymous No.10532102]

@10531972
[eqn]\sqrt2\;\in\;\mathbf R[/eqn]
[eqn]\sqrt2\, \left(\sqrt2\,+\,1\right)\, \left(\sqrt2\,+\,2\right)\, \left(\sqrt2\,+\,3\right) \,+\, 1 \;=\; 6\,\sqrt2 \,+\, 6 \,+\, 6 \,+\, 4 \,+\, 6\,\sqrt2 \,+\, 4\,\sqrt2 \,+\, 4\,\sqrt2 \,+\, 4 \,+\, 1 \;=\; 21 \,+\, 20\,\sqrt2[/eqn]
[math]21 \,+\, 20\,\sqrt2[/math] is not a square number.

[Anonymous No.10532108]

>>10532102
pic related

also:
>@

[Anonymous No.10532113]

Doesn't this have a solution where f(n)=1?

[Anonymous No.10532116]

>>10532113
meant -1

[Anonymous No.10532182]

>>10531972
[math]n^4 + 6 n^3 + 10 n^2 + 6 n + 1 = 0[/math]
[math]n^4 + 6 n^3 + 11 n^2 + 6 n + 1 = n^2[/math]

And why not a more obvious substitution of the presented problem.
[math]n(n+1)(n+2)(n+3)+1 = (n^2 + 3 n + 1)^2[/math]

*sigh* tum te tum. Next?