If a and b are 2 integers such that 7 divides a2+b2, show that 49 divides a2+b2
(Slight olympiad tilt here)
3a≡k mod(7) and b≡p mod(7)
we have a2+b2≡(k2+p2) mod 7
its given dat (k2+p2)=0 so k=p=0
so a,b shud be a multiple of 7
we got a2+b2=49(integer).
62sankara.. it is not given that k2+p2 = 0
but it is given that k2+p2 ≡ 0 (mod 7)
You have to slightly work on your argument from here...
11Squares are congruent to 1,4,2 modulo 7.
So that means 7|a and 7|b.
done.
341a^2 +b^2 \equiv 0 \pmod{7} \Rightarrow a^2 \equiv -b^2 \pmod 7
\Rightarrow a^6 \equiv -b^6 \pmod 7
If 7|x, then x6≡0 mod 7. Else by Fermat's theorem x6≡1 mod 7.
You can easily see now that we must have 7|a and 7|b for the above equation to be consistent.
This is true for any prime of the form 4k-1.
