prove, that if sqr(a) + sqr(b) divisible on 7, then a + b divisible on 7 too

Question

how can i prove, that if sqr(a) + sqr(b) divisible on 7, then a + b divisible on 7 too. i'm thinking on it whole day...

Thanks

update

i mean, that sqr(a) = a*a

Answer 1

Think about the residue of a number divided by 7, and the residue of its square divided by 7:

• 0 -> 0
• 1 -> 1
• 2 -> 4
• 3 -> 9 = 7 + 2
• 4 -> 16 = 14 + 2
• 5 -> 25 = 21 + 4
• 6 -> 36 = 35 + 1

So that the only possivle residues of squares are 0, 1, 2, 4. You also know that sum of two such numbers is divisible by 7. Hence the colcnlusion is: both numbers have residue 0, there's no other way to get something divisible by 7 from 0,1,2,4.

Conclusion: both numbers are divisible by 7. Of course this means that their sum is also divisible by 7.