Show that the square of any …
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Sia ? 6 years, 5 months ago
Let a be any positive integer.
Applying Euclid’s division lemma with divisor = 2, we get
{tex}\begin{array}{l}a=2q+r\;\;\;\;\;\;\;\;\;0\leq r<2\\So\;r=0,1\\\end{array}{/tex}
When r = 0,
a = 2q
So a2= (2q)2 = 4q2 = 4m--------(1) ( where m = q2, which is an integer)
When r = 1
Then a= 2q+1
a2= (2q + 1) 2 = 4q2 + 4q + 1 = 4(q2 + q ) +1 = 4m+1 --------(2) (where m = q2 + q, which is an integer)
From (1) and (2) We Can conclude that
The square of any positive integer is of the form 4m or 4m +1 for some integer m.
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