Prove that the product of four natural consecutive numbers is equal to the difference of the squares of two natural numbers.
$(a^2-d^2)n^4 + (2ab-2de)n^3 + (2ac+b^2-2df-e^2)n^2+(2bc-2ef)n+c^2-f^2 = \underbrace{(an^2+bn+c)^2-(dn^2+en+f)^2}_{\text{A difference of squares}} = n(n+1)(n+2)(n+3) = n^4+6n^3+11n^2+6n$
Equate coefficients on the far left with the far right:
Eq 1: $a^2-d^2 = 1$
Eq 2: $2ab-2de = 6$
Eq 3: $2ac+b^2-2df-e^2=11$
Eq 4: $2bc-2ef = 6$
Eq 5: $c^2-f^2 = 0$
Factoring Eq 1 gives:
$(a+d)(a-d) = 1$
The only factors of 1 are 1 and 1 or -1 and -1, so we have $a+d = a-d$. This gives us $d=0, a=\pm 1$. With no loss of generality, we can assume $a=1$. This process follows similarly for $a=-1$, but you will not wind up with a natural number for the first number. You will get negative the results you find for a,b,c. You can try it out yourself if you want. For now, assume $a=1$. Plug that in to Eq 2:
$2(1)b-2(0)e = 6$ implies $b=3$ and we are not sure about $e$. Let's plug in what we know to the last three equations:
Eq 3.1: $2(1)c+3^2-2(0)f-e^2=11$ implies $2c-e^2=2$. Since $-e^2 \le 0$, it must be that $2c \ge 2$ implies $c\ge 1$.
Eq 4.1: $2(3)c-2ef=6$ implies $3c-ef=3$. Again, we have $c\ge 1$.
Eq 5.1: $c^2-f^2=0$ implies $c=\pm f$.
Case 1: $c=f$
Plug into Eq 4.1:
$3c-ec=3$ implies $c(3-e)=3$. Therefore, $c=1$ or $c=3$ (note: we already figured out that $c$ must be positive from Eq 3.1). Then $3-e = 1$ or $3-e= 3$ are the only possibilities. Here are the possibilities:
$c=f=1; e=0$. Plugging into Eq 3.1 gives: $2(1)-0^2 = 2$, which is true. This gives: $(n^2+3n+1)^2-(0n^2+0n+1)^2 = (n^2+3n+1)^2-1$
$c=f=3; e=2$. Plugging into Eq 3.1 gives: $2(3)-2^2=2$, which is true. This gives $(n^2+3n+3)^2-(0n^2+2n+3)^2 = (n^2+3n+3)^2-(2n+3)^2$
Case 2: $c=-f$
Plug into Eq 4.1:
$3c+ec=3$ implies $c(3+e)=3$. Therefore, $c=1$ or $c=3$ and $3+e=3$ or $3+e=1$. This gives:
$c=1;f=-1;e=0$ which gives $(n^2+3n+1)^2-(0n^2+0n-1)^2$. But, the latter one is not a natural number, so we can throw this possibility out.
$c=3;f=-3;e=-2$ which gives $(n^2+3n+3)^2-(0n^2-2n-3)^2$. Again, this is never a natural number, so we can throw this possibility out, as well.
That gives us two possible differences of two squares of natural numbers:
$(n^2+3n+1)^2-1$
$(n^2+3n+3)^2-(2n+3)^2$
If we wanted, we could have used higher exponents than just $n^2$. But, that would make this far too complicated.
Note: . . . The content of my post was beaten in time by SlipEternal. I looked on my
page for somewhere to delete it so SlipEternal's post would be the last to show, but
I could not find any delete option.
Aside from changing all of the signs of the trinomial inside the parentheses, that is not the
only way. Here is another way:
$n(n + 1)(n + 2)(n + 3) = (n^2 + 3n + 3)^2 - (2n + 3)^2$
For instance, when n = 1, $ \ P(1) = 24 = 7^2 - 5^2.$
I looked at $ \ (an^2 + bn + c)^2 - (dn + f)^2 = n^4 + 6n^3 + 11n^2 + 6n = n(n + 1)(n + 2)(n + 3) \ \ $ **
I equated coefficients on the different degreed n variables and solved a system of equations.
There are many solutions.
Taking the non-negative integer solutions gives:
$(n^2 + 3n + 1)^2 - (1)^2$
$(n^2 + 3n + 3)^2 - (2n + 3)^2$
** I used "f" instead of "e" because of the other common use of e.
For a given $n$ let $p=n(n+1)(n+2)(n+3)$
For any divisor $r$ of $p$ , $r^2<p$, and both $r$ and $p/r$ even, let
$a=\frac{r}{2}+\frac{p}{2r}$ and
$b=\frac{p}{2r}-\frac{r}{2}$
We then have
$p=a^2-b^2$
so in general more than two solutions