Question

show that cube of any positive integer is of the form of4m,4m+14m+3 for some integer m

Answer 1

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

▶⏩By Euclid's Division lemma:-)

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

↪➡ a=4m+r. ,, 0≤ r< 4.

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

=> Taking r=0.

↪➡ a= 4m+0 → 4m.

=> Taking r=1.

↪➡ a= 4m+1.

=> Taking r=2.

↪➡ a= 4m+2.

=> Taking r=3.

↪➡ a= 4m+3.

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

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

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