UPSKILL MATH PLUS
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Learn moreProve that the square of any integer leaves the remainder either \(0\) or \(1\) when divided by \(4\).
Proof:
Let \(x\) be an integer.
The square of its integer is \(x^2\).
Here, we have two possibilities. \(x\) can be even or odd.
Case I: \(x\) be an even integer.
Let \(x = 4k\)
\(x^2 = (4k)^2 =\)\(k^2\)
\(x^2 = 4 \times \)\(k^2 + 0\)
The remainder \(r = 0\).
Case II: \(x\) be an odd integer.
Let \(x = 4k + 1\)
\(x^2 = (4k + 1)^2\)
\(x^2 = \)\((k^2) +\) \(k + 1\)
\(x^2 =\) \((k^2 + 2k) + 1\)
The remainder \(r = 1\).
We can conclude from the above cases that the square of any integer leaves a remainder of either \(0\) or \(1\) when divided by \(4\).
Hence, we proved.
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