Question

show that the square of any positive integer is either of the form for4m or 4 m + 1 for some integer m​

Answer 1

Answer:

let a be any positive interger and b=4, r=0,1,2,3,  b>r=>0

case (i) a=4q

a^2=(4q)^2=16q^2=4(4q^2)=4m

case (ii) a=4q+1

a^2=(4q+1)^2 = 16q^2+8q+1  =4(4q^2+2q) +1=4m+1

case (iii) a=4q+2

a^2=(4q+2)^2= 16q^2+16q+8=4(4q^2+4q+2)=4m

case (iv) a=4q+3

a^2=(4q+3)^2= 16q^2+24q+9=4(4q^2+6q+2)+1=4m+1

*HENCE VERIFIED

Answer 2

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