Question

prove that the square of any positive integer is of the form forum 4m + 1 for some integer m​

Answer 1

Let a be any +ve integer b=2 then by euclids division lemma a=bq+r such that 0 lessthan or equal to r less than b therefore the possible remainders are 0,1

a=(2q)^2=4q^2=4m                           (where m is some +ve integer)

a=(2q+1)^2 = 4q^2+4q+1=4m+1        (where m is some +ve integer)

Therefore the square of any +ve integer is of the form 4m or 4m+1

Hence proved