Show that the square of any …
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Ram Kushwah 6 years, 11 months ago
As per Euclid's Division Lemma
If a & b are 2 positive integers, then
a = bq + r {tex}0\;\leq b<r{/tex}
let b=3
hence r=0,1,2
If r=0
{tex}\begin{array}{l}a^2=(3q)^2=3x3q^2=3m\;(m=3q^2)--(1)\\if\;r=1\;then\;\\a^2=(3q+1{)^2=9q^2+6q+1=3}(3q^2+2q)+1=3m+1\;--(2)\;(m=3q^2+2q)\;---(2)\\if\;r=2\\a^2=(3q+2)^2=9q^2+12q+4=3(3q^2+4q+1)+1=3m+1--(3)\;\;(m=3q^2+4q+1)\end{array}{/tex}
From (1) ,(2) and (3)
square of any positive integer a is in the form of 3m or 3m +1
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