Question

show that the square of any positive integer cannot be of the 5p+2 or 5p+3 for any integer p​

Answer 1

Answer: I hope it may help you

Step-by-step explanation:

Let a be any positive integer and b =5.

Using Euclid's Division Lemma,we have

a = bq +r , 0 ≤ r∠b

a = 5q+r , where r =0,1,2,3 and 4.

so, a can be =5q,5q+1,5q+2,5q+3 and 5q+4.

taking a = 5q

on squaring both sides, we get

a² = 25q²

a² = 5(5q²)

a² = 5p    { let 5q² = p for some integer p}

taking a = 5q+1

on squaring both sides, we get

a² = (5q+1)²

a² = (5q)²+2×5q×1 + 1²

a² = 25q² + 10q + 1

a² = 5(5q² + 2) + 1

a² = 5p + 1    (let 5q²+2 =p for some integer p)

then taking a = 5q+2  

squaring both sides ,

a² = (5q+2)²

a² = 25q² + 20q + 4

a² =  5(5q² + 4q ) + 4

a² = 5p + 4  (let 5q²+4q = p for some integer p)

Now ,taking a = 5q+3

On squaring both sides , we get

a² = 25q²+30q + 9

a² = 5 (5q² + 6q + 1) +4

a² = 5p + 4 (let 5q² + 6q + 1 = p for some integer p )

From the above,the square of any positive integer can not be 5p +2 or 5p +  3. It will be of the form 5p,5p + 1 and 5p+4.

                                                                       

                                                             Hence Proved.