Question

show that square of any positive integer is of the form
7p and 7p +1 and 7p+4 for some integer p ( explaintion plz it's urgent)​ with explaintion​

Answer 1

$\large\bf\underline{Given:-}$

• p is some intiger

$\large\bf\underline {To \: find:-}$

• show that square of any positive integer is of the form 7p and 7p +1 and 7p+4 for some integer p.

$\huge\bf\underline{Solution:-}$

Let b be the divisor and q be the quotient and r = remainder.

»★Let the positive intiger be a

• b = 7

≫By Euclid Division lemma:-

• a = bq + r
• 0≤r<b

✰ a = 7q + r

• Where, r = 0, 1 , 2 ,3 ,4 ,5 ,6

$\underbrace{ \bf \: case \: 1st \: : - }$

»» a = 7q + 0

• Squaring on both sides

»» a² = 49q²

»» a² = 7(7q²)

»» a² = 7p⠀⠀⠀⠀⠀⠀ [p = 7q²]

$\underbrace{ \bf \: case \: 2nd \: : - }$

»» a = 7q + 1

• Squaring on both sides

»» a² = (7q+1)²

»» a² = 49q² + 1 + 14q

»» a² = 7(7q² +2q)+1

»» a² = 7p+1⠀⠀⠀⠀⠀⠀[p = 7q² + 2q]

$\underbrace{ \bf \: case \: 3rd \: : - }$

»» a = 7q + 2

• Squaring on both sides

»» a² = (7q +2) ²

»» a² = 49q² + 4 + 28q

»» a² = 7(7q² +4q)+4

»» a² = 7p + 4⠀⠀⠀⠀⠀⠀[p = 7q² + 4q]

So,

The square of any positive integer is of the form 7p and 7p +1 and 7p+4 for some integer p

Answer 2

$\large\mathrm\red{Given:-}$

• p is some integer

• show that square of any positive integer is of the form 7p and 7p +1 and 7p+4 for some integer p.

Let b be the divisor and q be the quotient and r = remainder.

»★$\large\mathrm\red{Let \: the \: positive \: intiger \: be \: a}$

$\implies$b = 7

≫$\large\mathrm\red{By \: Euclid \: Division \: lemma:-}$

a = bq + r

0≤r<b

✰ a = 7q + r

$\large\mathrm\red{Where}$, r = 0, 1 , 2 ,3 ,4 ,5 ,6

»» a = 7q + 0

$\large\mathrm\red{1. \: When \: r \: value \: equals \: to \: zero. \: (r = 0)}$

»» a = 7q + r

»» a = 7q

$\large\mathrm\red{Cube \: both \: side}$

»» a³ = 343³

»» a³ = 7(49q³)

»» a³ = 7m. (/here m is positive integer and m = 49q³)

$\large\mathrm\red{2. \: If \: value \: of \: r \: is \: equal \: to \: 1 \: (r = 1)}$

»» a = 7q + r

»» a = 7q + 1

$\large\mathrm\red{Cube \: both \: side}$

»» a³ =(7q + 1)³

»» a³ = 343q³ + 1³ + 3(7q)(1)(7q + 1)

»» a³ = 343q³ + 21q(7q + 1) + 1

»» a³ = 343q³ + 147q² + 21q + 1

»» a³ = 7m + 1. (/here m is positive integer and m = 49q³ + 21q² + 3q)

$\large\mathrm\red{3. \: If \: r \: value \: equal \: to \: 6 \: (r = 6)}$

»» a = 7q + r

»» a = 7q + 6

$\large\mathrm\red{Cube \: both \: side}$

»» a³ = (7q + 6)³

»» a³ = 343q² + 216 + 3(7q)(6)(7q + 6)

»» a³ = 343q³ + 126q(7q + 6) + 216

»» a³ = 7(49q³ + 127q² + 108q) + 216

»» a³ = 7m + 216. (/here m is positive integer and m = 49q³ + 127q² + 108q)

$\large\mathrm{I \: think \: 7m + 6 \: is \: not \: a \: cube \: of \: any \: position \: integer.</p><p>}$

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