If $p$ is an integer, then every square integer is of the form
- $2p$ or $(4p-1)$
- $4p$ or $(4p-1)$
- $3p$ or $(3p+1)$
- $4p$ or $(4p+1)$
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Correct answer
Explanation
Let $x$ be any positive integer.
For example Let $x=2\Rightarrow\,{x}^{2}=4$ is a square number is of the form $4p$ where $p\in Z$
Let $x=3\Rightarrow\,{x}^{2}=9=2\times 4+1$ is a square number is of the form $4p+1$ where $p\in Z$
Let $x=4\Rightarrow\,{x}^{2}=16$ is a square number is of the form $4p$ where $p\in Z$
Let $x=5\Rightarrow\,{x}^{2}=25=4\times 6+1$ is a square number is of the form $4p+1$ where $p\in Z$
Let $x=4p\Rightarrow\,{x}^{2}=16{p}^{2}=4\left(4{p}^{2}\right)=4q$ where $q=4{p}^{2}$ is a square number is of the form $4p$ where $p\in Z$
If $x=4p+1\Rightarrow\,{x}^{2}={\left(4p+1\right)}^{2}=16{p}^{2}+8p+1=4\left(4{p}^{2}+2p\right)+1=4q+1$ where $q=4{p}^{2}+2p$
If $x=4p+3\Rightarrow\,{x}^{2}={\left(4p+3\right)}^{2}=16{p}^{2}+24p+9=4\left(4{p}^{2}+6p+2\right)+1=4q+1$ where $q=4{p}^{2}+6p+2$
$\therefore\,$in each of the above cases,${x}^{2}$ is of the form $4p$ or $4p+1$