# Help with question

Thread starter 2 years ago
#1
An integer N is the sum of the squares of two different integers.
i) Prove that N^2 is also the sum of the squares of two integers.

So far I have tried:
N = a^2 + b^2
N^2 = (a^2 + b^2)^2 = a^4 + b^4 + a^2b^2
0
2 years ago
#2
(Original post by Cpt Avocado)
An integer N is the sum of the squares of two different integers.
i) Prove that N^2 is also the sum of the squares of two integers.

So far I have tried:
N = a^2 + b^2
N^2 = (a^2 + b^2)^2 = a^4 + b^4 + a^2b^2
You should have two lots of a^2b^2, not just one.

I'm sure you'll agree that 2a^2b^2 = 4a^2b^2 - 2a^2b^2. Can you see how this helps you?
0
2 years ago
#3
(Original post by Cpt Avocado)
An integer N is the sum of the squares of two different integers.
i) Prove that N^2 is also the sum of the squares of two integers.

So far I have tried:
N = a^2 + b^2
N^2 = (a^2 + b^2)^2 = a^4 + b^4 + a^2b^2
Should be

N^2 = a^4 + b^4 + 2a^2b^2

Anyway, rewrite as

(a^4 - 2a^2b^2 + b^4) + 4a^2b^2

and finish off from there.
0
Thread starter 2 years ago
#4
(Original post by Pangol)
You should have two lots of a^2b^2, not just one.

I'm sure you'll agree that 2a^2b^2 = 4a^2b^2 - 2a^2b^2. Can you see how this helps you?
Yes that was a typo. I'll have another go with that hint
0
Thread starter 2 years ago
#5
So it's (a^2 - b^2)^2 + (2ab)^2

My brain just doesn't make that leap
Thanks tho!
0
2 years ago
#6
(Original post by Cpt Avocado)
So it's (a^2 - b^2)^2 + (2ab)^2

My brain just doesn't make that leap
Thanks tho!
It's hard to know what you've got to do when you've never seen a problem like this before. So much of knowing what to do is down to experience. Try and remember this trick for the future!
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