Question

4. Prove that square of a positive integer is of the form 4m or 4m+1 for some
positive integer m.​

Answer 1

☞ Let a be any positive integer and b = 4

Then by using Euclid's algorithm

a = 4q + r __________(eq 1)

Here, r = 0, 1, 2, 3 ..

On putting the above values of r in (eq 1) we get;

a = 4q, 4q + 1, 4q + 2 and 4q + 3

• Since, a is an odd integer.

So, we take a = 4q, 4q + 1and 4q + 3.

_______________________________

$\underline{Case\:1.}$

a = 4q

• Squaring on both sides.

=> a² = (4q)²

=> 16q²

=> 4(4q²)

[Take 4q² = m]

=> 4m

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$\underline{Case\:2.}$

If a = 4q + 1

• Squaring both sides

=> a² = (4q + 1)²

=> 16q² + 1 + 8q

=> 4(4q² + 2q) + 1

[Take m = 4q² + 2q]

=> 4m + 1

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$\underline{Case\:3.}$

Ifa = 4q + 3

• Squaring both sides, we have,

=> a² = (4q +3)²

=> 16q² + 9 + 24q

=> 16q² + 24q + 8 + 1

=> 4(4q² + 6q + 2) +1

[Take m = 4q² + 7q + 2]

=> 4m +1

_______________________________

Hence, a is of the form 4m or 4m + 1 for some integer m.

___________ [HENCE PROVED]