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Show that any positive odd integer …

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Show that any positive odd integer is in form of 4m+1 or 4m+3

Posted by Prajjwal Kapri 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let a be the positive integer and b = 4.
Then, by Euclid’s algorithm, a = 4q + r for some integer q ≥ 0 and r = 0, 1, 2, 3 because 0 ≤ r < 4.
So, a = 4q or 4q + 1 or 4q + 2 or 4q + 3.
{tex}(4q)^3\;=\;64q^3\;=\;4(16q^3){/tex}
= 4m, where m is some integer.
{tex}(4q+1)^3\;=\;64q^3+48q^2+12q+1=4(\;16q^3+12q^2+3q)+1{/tex}
= 4m + 1, where m is some integer.
{tex}(4q+2)^3\;=\;64q^3+96q^2+48q+8=4(\;16q^3+24q^2+12q+2){/tex}
= 4m, where m is some integer.
{tex}(4q+3)^3\;=\;64q^3+144q^2+108q+27{/tex}

=4×(16q³+36q²+27q+6)+3
= 4m + 3, where m is some integer.
Hence, The cube of any positive integer is of the form 4m, 4m + 1 or 4m + 3 for some integer m.

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