by Euclid alogarithm prove that square of any positive integer is of the form 3p,3p+1
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r = 0,1,2
integer be a=bq+r
r=0
a = (3q+0)² = 9q² = 3p where p = 3q²
a = (3q+1)² = 9q² +1 + 6q = 3p+1 where p = 3q² +2q
a= (3q+2)² = 9q² + 4 + 12q = 3p+1 where p = 3q² +1 +4q
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integer be a=bq+r
r=0
a = (3q+0)² = 9q² = 3p where p = 3q²
a = (3q+1)² = 9q² +1 + 6q = 3p+1 where p = 3q² +2q
a= (3q+2)² = 9q² + 4 + 12q = 3p+1 where p = 3q² +1 +4q
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akularohan265:
if we take r=3, we get 9(p^2+2p)+1.it is not of the form 3p,3p+1
sorry I got it
ok
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