<s>[INST] Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$ Find $n$ such that $(m-1)(m+3)(m-4)(m-8)+n$ is a perfect square for all $m$

I am thinking of starting like this
$(m-1)(m+3)(m-4)(m-8)+n = k^2 \implies (m-1)(m+3)(m-4)(m-8) = k^2-n$
Honestly somewhat scared of expanding the products on left hand side. Any hints/help on the next step ? Thanks! 
 [/INST] The trick is to do $(m-1)(m+3)(m-4)(m-8)=\underbrace{(m-1)(m-4)}\underbrace{(m+3)(m-8)}$ since $-1-4=-5=3-8$ giving same coefficient of m:

$$(m-1)(m+3)(m-4)(m-8)+n=(m^2-5m+4)(m^2-5m-24)+n\stackrel{y=m^2-5m+4}=y(y-28)+n=y^2-28y+n=(y-14)^2+n-14^2\implies n=14^2=196$$
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