Question

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

Answer 1

✴Hey friends!!✴✴

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✴✴ Here is your answer↓⬇⏬⤵

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=> Let a be the any positive integer.
:-) then b = 4.

By Euclid's Division lemma:-)

↪➡ a= bq+r. [ q = Quotient].

↪➡ 0≤r<b.
=> 0≤r<4.

:-( r= 0,1,2,3).

⏩▶ Taking r = 0.

=> a= bq+r.

↪➡ a= 4q+0.

↪➡ a=4q.

↪➡a=(4q)².

↪➡ a=16q².

↪➡ a= 4(4q)².

↪➡a= 4m. [ where m = 4q²].

Now,

⏩▶ Taking r= 1.

=> a= bq+r.

↪➡ a= 4q+1.

↪➡ a= (4q+1)².

↪➡ a= 16q²+ 8q+1.

↪➡ a= 4(4q²+2q)+1.

↪➡ a= 4m+1. [ where m= 4q²+2q].

✴✴ Hence, it is proved that 4q and 4q+1 is the any positive integers for some integers q.✴✴✔✔.

✴✴ Thanks ✴✴.

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Answer 2

☆Hey friend!!!! ☆

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here is your answer ☞
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let 4q and (4q+1) be any positive integers

=> x = 4q



squaring both sides 

=> x² = 16q²

=> x²= 4(4q²)

put 4q² be m 

=> x = 4m

also 4q+1 =x

squaring both sides

=> x² = 16q² +1 +8q 

=> x = 4(4q² + 2q) +1

put 4q² + 2q = m

=> x = 4m +1

Hence proved

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