Proof If 7|(a^2+b^2) then 7|a and 7|b

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SUMMARY

The discussion focuses on proving the statement: If 7 divides \(a^2 + b^2\), then 7 divides both \(a\) and \(b\). Participants analyze the periodicity of quadratic residues modulo 7, identifying the residues as {0, 1, 2, 4}. The conclusion drawn is that the sum of two quadratic residues can only equal zero modulo 7 if both residues are zero, thereby confirming that if \(7 | (a^2 + b^2)\), then \(7 | a\) and \(7 | b\).

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lukaszh
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Hello,
how to prove
If 7|(a^2+b^2) then 7|a and 7|b.
(If seven divides a^2+b^2 then seven divides a and seven divides b)
Thanks.
 
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Hello lukaszh! :smile:

(try using the X2 tag just above the Reply box :wink:)
lukaszh said:
Hello,
how to prove
If 7|(a^2+b^2) then 7|a and 7|b.
(If seven divides a^2+b^2 then seven divides a and seven divides b)
Thanks.

Hint: what are 12, 22, 32 etc (mod 7)? :wink:
 
12=1 (mod7)
22=4 (mod7)
32=2 (mod7)
42=2 (mod7)
52=4 (mod7)
62=1 (mod7)
72=0 (mod7)
Is it periodic {1,4,2,2,4,1,0} ? Now I know :-) It's periodic, so if I add any of these congruences together there will be some remainder. Remainder is zero if and only if I add congruences in form
(7k)2=0 (mod7)
(7j)2=0 (mod7)
THANX :-)
 

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