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Show that for odd positive integer to be a perfect square, it should be of the form
8k +1. Let a=2m+1
Ans: Squaring both sides we get a2 = 4m (m +1) + 1
∴ product of two consecutive numbers is always even
m(m+1)=2k
a2=4(2k)+1 a2 = 8 k + 1
Hence proved
the segments shown could form a triangle
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