Question

show that the square of any positive integer can't be of the form 5m+2or 5m+3 where 'm' is a whole number ​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

TO PROVE

The square of any positive integer can't be of the form 5m+2or 5m+3 where m is a whole number

PROOF

In order to prove the given result we need to memorise THE DIVISION ALGORITHM, which states that :

If a & b are integers where b is a positive integer then there exist unique integers q & r such that

$\sf{ \: a = bq + r \: \: \: \: \: with \: 0 \leqslant r < b \: }$

Here b = 5

So

$\sf{ 0 \leqslant r < 5\: }$

$\sf{\implies \: r = 0, 1,2,3,4\: }$

$\sf{ \underline{CASE : 1 \: ( r = 0)}}$

$\sf{a = 5q}$

$\therefore \: \sf{ { \: {a}^{2} = {(5q )}^{2} \: }}$

$\: \sf{ { \: = 25 {q}^{2} \: }}$

$\: \sf{ { \: = 5 \times 5 {q}^{2} \: }}$

$\: \sf{ { \: = 5m \: }} \: \: \: where \: \: m = 5 {q}^{2}$

$\sf{ \underline{CASE : 2 \: \: ( r = 1)}}$

$\sf{a = 5q + 1}$

$\therefore \: \sf{ { \: {a}^{2} = {(5q + 1)}^{2} \: }}$

$\: \sf{ { \: = 25 {q}^{2} + 10q + 1 \: }}$

$\: \sf{ { \: = 5(5 {q}^{2} + 2q) + 1 \: }}$

$\: \sf{ { \: = 5m + 1 \: }} \: \: where \: \: m = 5 {q}^{2} + 2q$

$\sf{ \underline{CASE : 3 \: \: ( r = 2)}}$

$\sf{a = 5q + 2}$

$\therefore \: \sf{ { \: {a}^{2} = {(5q + 2)}^{2} \: }}$

$\: \sf{ { \: = 25 {q}^{2} + 20q + 4 \: }}$

$\: \sf{ { \: = 5m+ 4 \: } \: \: \: where \: \: \: m = 5 {q}^{2} + 4q \: }$

$\sf{ \underline{CASE : 4 \: \: ( r = 3)}}$

$\sf{a = 5q + 3}$

$\therefore \: \sf{ { \: {a}^{2} = {(5q + 3)}^{2} \: }}$

$\: \sf{ { \: = 25 {q}^{2} + 30q + 9 \: }}$

$\: \sf{ { \: = 5(5 {q}^{2} + 6q +1) + 4 \: }}$

$\: \sf{ { \: = 5m + 4 \: }} \: \: where \: \: m = 5 {q}^{2} + 6q +1$

$\sf{ \underline{CASE : 5 \: \: ( r = 4)}}$

$\sf{a = 5q + 4}$

$\therefore \: \sf{ { \: {a}^{2} = {(5q + 4)}^{2} \: }}$

$\: \sf{ { \: = 25 {q}^{2} + 40q + 16 \: }}$

$\: \sf{ { \: = 5 (5{q}^{2} + 8q + 3) + 1 \: }}$

$\: \sf{ { \: = 5m+ 1 \: }} \: \: where \: \: m = 5{q}^{2} + 8q + 3$

So From above we visualize that the square of any positive integer can't be of the form 5m+2or 5m+3 where m is a whole number

Hence the proof follows

Answer 2

n order to prove the given result we need to memorise THE DIVISION ALGORITHM, which states that :

If a & b are integers where b is a positive integer then there exist unique integers q & r such that