Help with proof q (1 Viewer)

[Run hard@thehsc]

 

[amdspotter]

View attachment 34577

Sup did this one somewhere b4 but don't have physical sol rn so bro but do this like factor a out and then simplify the quadratic U have
Now U get three terms
A-1 a and a+1
From here U know these are consecutive
So like
Hence one of a-1 and a must be a multiple of 2
And in
A-1 a and a+1 one of them must be multiple of 3
Voila
This means a-1 a and a+1 are multiples of both 2 and 3

And hence
A3-a is a divisible by 6 cuz it's multiple of 6

Sos if it's confusing the way I've formatted it. I'm pretty sure this is the way to do it the q

 

[Run hard@thehsc]

Cool so I did do it right! Thanks!

 

[5uckerberg]

View attachment 34577

This question is a simple matter of factorising which in turn gives us going from for something to be divisible by 6 the factors when added have to be divisible by 3 and it has to be even. To prove that this statement is even it has been implied that for all a which are integers regardless of what value of a is it will give a even number and two odd numbers or two even numbers and an odd number and when multiplied will still be even. Note with the addition of factors you have clearly that is divisible by 3 nd combining with the fact it is even we can conclude is divisible by 6.