Proofs q (1 Viewer)

[Run hard@thehsc]

Need help with 'c' - how do you do this?

 

[Trebla]

From part b, the sum of the squares of two odd numbers is 8(M+N) + 2 for some positive integers M and N.

This is an even number. In order for it to be a square number, a necessary condition (though not sufficient) is that its square root must be even which implies it must be divisible by 4. Clearly it is not possible to factorise 4 in that expression and be left with an integer.

 

[=)(=]

Try this ( someone please correct me if i am wrong)

let a=2n+1 and b=2n-1

squaring the a and squaring b and finding their sum results in

4n^2 +4n +1 +4n^2 -4n +1
which results in 8n^2

Now to be a square the square root of the result has to be an integer
sqrt(8n^2) can be broken into the sqrt(8)x sqrt(n^2)
which equals but since root 2 is irrational then is not a square

 

[Run hard@thehsc]

That makes sense - thanks everyone