Show that the square of an odd integer is of the form 4q+1, for some integer q.
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Show that the square on any odd integer is of the form (4q+1) for some integer q.
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By Euclid's division algorithm , a=bq+r where a,b,q,r are non-negative integers and 0≤r<b.
On putting b=4 we get
a=4q+r ....(i)
When r=0,a=4q which is even (as it is divisible by 2)
When r=1,a=4q+1 which is odd (as it is not divisible by 2)
Squaring the odd number (4q+1) , we get
a
2
=(4q+1)
2
=(4q)
2
+(1)
2
+2(4q)
=4[4q
2
+2q]+1
=4m+1 is perfect square for m=4q
2
+2q
When r=2,a=4q+2 [From(i)]
⇒a=2(2q+1) which is even (as it is divisible by 2)
When r=3,a=4q+3=4q+2+1
=2[2q+1]+1 which is odd (as it is not divisible by 2)
Squaring the odd number (4q+3) , we get
a
2
=(4q+3)
2
=(4q)
2
+(3)
2
+2(4q)(3)
=16q
2
+9+24q
=16q
2
+24q+8+1
=4[4q
2
6q+2]+1
=4m+1 is perfect square for some value of m.
Hence, the square on any odd integer is of the form (4q+1) for some integer q.
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