A number theory problem by A Former Brilliant Member

substitute $m^2+3m=x$ .Then, x(x+2)=n(n+1).

Here, if x>n then LHS >RHS.n=x doesn't work.Again if n>x then n+1>x+1 and consequently n+1 $\ge x+2$ ,which implies RHS>LHS.

So,either way we get a contradiction.

Consider the numbers $m(m+3)$ and $(m+1)(m+2)$ , which multiply to give $m(m+1)(m+2)(m+3)$ . For example, $6\times7\times8\times9=3024=(6\times9)(7\times8)=54\times56$ Notice that the difference between these two numbers is $2$ . This is in fact true for any value of $m$ : $(m+1)(m+2)-m(m+3)=m^2+3m+2-m^2-3m=2$ Call these two numbers $n$ and $k$ . If we want $nk$ to equal a fixed number, in this case $m(m+1)(m+2)(m+3)$ , then if we increase $n$ we must decrease $k$ , and vice versa.

We currently have $k=n+2$ , So, in order to have $k=n+1$ , we have to increase $n$ and decrease $k$ . But, there is no way to do this while keeping $n$ and $k$ as integers and having $n<k$ , so it is not possible to achieve the required result.

Hence, given a positive integer $m$ it is impossible to find a positive integer $n$ so that $m(m+1)(m+2)(m+3)=n(n+1)$ .

$m(m+1)(m+2)(m+3)=n(n+1)$

$(m^2+m)(m^2+5m+6)=n^2+n$

$m^4+5m^3+6m^2+m^3+5m^2+6m=n^2+n$

Add one to each side:

$m^4+6m^3+11m^2+6m+1=n^2+n+1$

Formulas: $a^2+2ab+b^2=(a+b)^2$ and $a(b)+c(b)=b(a+c)$

$[m^4+6m^3+9m^2]+[2m^2+6m]+1=n^2+n+1$

$[(m^2+3m)^2+2(m^2+3m)]+1=n^2+n+1$

$(m^2+3m)(m^2+3m+[2])+1=n^2+n+1$

$(m^2+3m)(m^2+3m+1)+m^2+3m+1=n^2+n+1$

$(m^2+3m+1)(m^2+3m+1)=n^2+n+1$

Now we know that:

$n^2< n^2+n+1 <n^2+2n+1$

$n^2< n^2+n+1 <(n+1)^2$

So:

$n^2< (m^2+3m+1)^2 <(n+1)^2$

But in this interval there is no integer square.