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

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To prove that if 7 divides a² + b², then 7 must also divide both a and b, one can analyze the quadratic residues modulo 7. The possible residues for squares modulo 7 are 0, 1, 2, and 4, forming a periodic pattern. The only way for the sum a² + b² to equal 0 modulo 7 is if both a² and b² are 0 modulo 7, which implies that both a and b must be divisible by 7. Thus, the proof relies on the periodicity of the quadratic residues and their sums. This establishes the necessary condition that if 7 divides a² + b², then it must divide both a and b.
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 :-)
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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