Proving any product of four consecutive integers is one less than a perfect square Prove or disprove that : Any product of four consecutive integers is one less than a
perfect square.
OK so I start with $n(n+1)(n+2)(n+3)$ which can be rewritten $n(n+3)(n+1)(n+2)$
After multiplying we get $(n^2 + 3n)(n^2 + 3n + 2)$
How do I proceed from here to end up with something squared $- 1$?
 A: Here's another approach you might use, a sort of brute-force approach.  You said you got as far as $$(n^2 + 3n)(n^2 + 3n + 2).$$  Continuing to multiply, you get: $$ n^4 + 6n^3 + 11n^2 + 6n.$$
The claim is that that this is one less than a perfect square, or equivalently, that $$n^4 + 6n^3 + 11n^2 + 6n + 1$$ is a perfect square.  If this last expression is a perfect square, it must be the square of something of the form $n^2 + an+b$ for some $a$ and $b$.  But what are $a$ and $b$?
Squaring $n^2 + an+b$, we get $$n^4 + 2an^3 +(a^2+2b)n^2 + (2ab)n + b^2$$
and equating the coefficients of the two polynomials we get $$\begin{align}
2a & = 6 \\
a^2+2b & = 11\\
2ab & = 6 \\
b^2 & = 1
\end{align}$$
which we can easily solve to obtain $a=3, b=1$.  So putting together the various steps we have $$n(n+1)(n+2)(n+3) = (n^2 + 3n + 1)^2 - 1$$ which is what we were looking for.
A: Observe that the product $p$ of four consecutive integers can be written as $p=(x-\frac{3}{2})(x-\frac{1}{2})(x+\frac{1}{2})(x+\frac{3}{2})$ where $x=n+\frac{1}{2}$ for some integer $n$. Then $p=(x^2-\frac{9}{4})(x^2-\frac{1}{4}) = (x^2-\frac{5}{4}+1)(x^2-\frac{5}{4}-1) = (x^2-\frac{5}{4})^2-1$. It remains to show that $x^2-\frac{5}{4}$ is an integer, which is an easy bit of arithmetic.
A: Well you can write the equation as this
$$(n^2+3n)(n^2+3n+2)=(n^2+3n+1-1)(n^2+3n+1+1)=(n^2+3n+1)^2-1^2$$
If you know you have to make this equation into an perfect square you need to get 2 same things and -1
A: You might note that for any $N$ at all, $$(N-1)(N+1) = N^2-1$$ and so is one less than a perfect square. I presume you have seen this before.
Then you could take $N = M+1$ in the formula above, and get $$M\cdot(M+2) = (M+1)^2 - 1$$ is one less than a perfect square.
And here you have $M = n^2+3n$, don't you?
A: let the nos be 
n-1, n, n+1, n+2
we need to prove that (n-1)(n)(n+1)(n+2)+1= square no.
rearranging it : (n-1)(n+2)(n)(n+1)+1= square no.
LHS= (n-1)(n+2)(n)(n+1)+1
   =(n^2+n-2)(n^2 +n)+1
   = (k-2)(k) + 1                    (by replacing (n^2 +n) by k)
   =  k^2 -2k +1
   = (k-1)^2 
 hence proved , LHS is a square no.
on an interesting note,
(k-1)^2 = (n^2 +n-1)^2= {(n)(n+1)-1} ^2
therefore, the no. formed is always the square of, product of the two middle terms -1.  
A: It seems like the way to attack this that doesn't require guessing is this:  Start with $N (N-1) (N-2) (N-3) = N^4 + 6 N^3 + 11 N^2 + 6 N = M^2 - 1$  Since the product is "near" $N^4$, $M$ has to be "near" $N^2$.  And it's very likely that $M$ is a polynomial in $N$.  So set $M = N^2 + aN + b$.  Then $M^2-1 = N^4 + 2aN^3 + (2b+a^2)N^2 + 2abN + (b^2-1)$.  Setting that into the first equation and equating powers of $N$ gives $2a = 6, 2b+a^2 = 11, 2ab = 6, b^2-1 = 0$, which has the unique solution $a = 3, b = 1$.  So $M = N^2 + 3N + 1$ and $N (N-1) (N-2) (N-3) = (N^2 + 3N + 1)^2 -1$.
