Question

show that the square of any positive integer is of the form 4q+1 , or 4q+3 where q is some integers

Answer 1

Let a be any positive integer and b =3.By division Lemma there exits integers q and r such that. a=4q+r,where 0 <r <4 => a=4q or ,a=4q+1 or a=4q+2 or 4q+3 {0 <r <4=>r=0,1,2,3} =>a=4q +1 or ,a=4q+3( a is positive integer a≠ 4q,a≠4q+2) Hence,any positive integer is of the form 4q+1 or 4q+3

Answer 2

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▶⏩ Let ‘a’ be the any positive integers,
=> then, b= 4.

▶⏩$\boxed{By \: Euclid's \: Division \: lemma:-)}$

↪➡ a= bq+r. ,, 0≤ r< b. [ m= quotient].

↪➡ a=4q+r. ,, 0≤ r< 4.

→ Hence, possible values of r= 0,1,2,3.

=> Taking r=0.

↪➡ a= 4q+0 → 4q.

=> Taking r=1.

↪➡ a= 4q+1.

=> Taking r=2.

↪➡ a= 4q+2.

=> Taking r=3.

↪➡ a= 4q+3.

▶⏩ Hence, some odd integers are 4q+1 or 4q+3.

✴✴ Therefore, it is proved that 4q+1 and 4q+3 are positive odd integers for some integers m. ✴✴.

✴✴$\boxed{THANKS}$✴✴

☺☺☺$\bf \underline{Hope \: it \: is \: helpful \: for \: you}$✌✌✌.