then xy is divisible by 7
show that the condition (x,y)=1 is necessary/
(x,y) means their gcd.
divisibility
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prove that if x,y,z are positive integers such that (x,y)=1 and x^2+y^2=z^4
then xy is divisible by 7
show that the condition (x,y)=1 is necessary/
(x,y) means their gcd.
then xy is divisible by 7
show that the condition (x,y)=1 is necessary/
(x,y) means their gcd.
women of purity are for men of purity and hence men of purity are for women of purity - THE HOLY QURAN
Re: divisibility
Let $z^2=m^2+n^2,x=2mn,y=m^2-n^2$ where $gcd(m,n)=1$
Then again $z=u^2+v^2,m=u^2-v^2,n=2uv$
Setting $xy=2mn(m+n)(m^2-n^2-2mn)(m^2-n^2+2mn),$note that all the $5$ factors except $2$ are pairwisely coprime,if no number divisible by $7,$then from box principle,one must be divisible by $7,$that is $7|xy$
Then again $z=u^2+v^2,m=u^2-v^2,n=2uv$
Setting $xy=2mn(m+n)(m^2-n^2-2mn)(m^2-n^2+2mn),$note that all the $5$ factors except $2$ are pairwisely coprime,if no number divisible by $7,$then from box principle,one must be divisible by $7,$that is $7|xy$
One one thing is neutral in the universe, that is $0$.