show that square of any positive integer cannot be of the from 5q+2 or 5q+3 for any integer q
Answers
Answer:
Let a be the positive integer and b = 5.
Then, by Euclid’s algorithm, a = 5m + r for some integer m ≥ 0 and r = 0, 1, 2, 3, 4 because 0 ≤ r < 5.
So, a = 5m or 5m + 1 or 5m + 2 or 5m + 3 or 5m + 4.
So, (5m)2 = 25m2 = 5(5m2) = 5q, where q is any integer.
(5m + 1)2 = 25m2 + 10m + 1 = 5(5m2 + 2m) + 1 = 5q + 1, where q is any integer.
(5m + 2)2 = 25m2 + 20m + 4 = 5(5m2 + 4m) + 4 = 5q + 4, where q is any integer.
(5m + 3)2 = 25m2 + 30m + 9 = 5(5m2 + 6m + 1) + 4 = 5q + 4, where q is any integer.
(5m + 4)2 = 25m2 + 40m + 16 = 5(5m2 + 8m + 3) + 1 = 5q + 1, where q is any integer.
Hence, The square of any positive integer is of the form 5q, 5q + 1, 5q + 4 and cannot be of the form 5q + 2 or 5q + 3 for any integer q.
Answer: using euclid's division lemma
that is "a=bq+r"
let a be any positive integer
let it be divided by 5
and quotient be m
therefore a=5m+r .......................... where r can be 0 ≤ r > 5
possible forms of a can be
case (i) a=5m
case (ii) a=5m+1
case (iii) a=5m+2
case (iv) a=5m+3
case (v) a=5m+4
case (i)
a=5m
a²=(5m)²
a²=25m²
a²=5(5m²)
on considering (5m²)=q
then , a²= 5q
similarly we solve case(ii), case(iii), case(iv), case(v)
after solving we get
case(i) =5q
case(ii)=5q+1
case(iii)=5q+4
case(iv)=5q+4
case(v)=5q+1
hence we can say that the square of any positive
integer cannot be in the form of 5q+2 or 5q+3
..................hope this my help you..................................