prove that square of any positive integer is of the form 4q or 4q+1 for some integer q
Answers
Answer:
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Step-by-step explanation:
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
Squaring both sides, we get
a
2
=(4q)
2
=4(4q
2
)
=4m is perfect square for m=4q
2
When r=1,a=4q+1
Squaring both sides, 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
Squaring both sides, we get
a
2
=(4q+2)
2
=(4q)
2
+(2)
2
+2(4q)(2)
=4(4q
2
+4q+1)
=4m is perfect square for m=4q
2
+4q+1
When r=3,a=4q+3
Squaring both sides, 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 integer is either of the form 4q or (4q+1) for some integer q.