Show that the square of any …
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Sia ? 6 years, 4 months ago
Let a = 4q + r, when r = 0, 1, 2 and 3
{tex}\therefore{/tex}Numbers are 4q, 4q + 1, 4q + 2 and 4q + 3
{tex}( a ) ^ { 2 } = ( 4 q ) ^ { 2 } = 16 q ^ { 2 } = 4 ( 4 q ) ^ { 2 } = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 1 ) ^ { 2 } = 16 q ^ { 2 } + 8 q + 1 = 4 \left( 4 q ^ { 2 } + 2 q \right) + 1 = 4 m + 1{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 2 ) ^ { 2 } = 16 q ^ { 2 } + 16 q + 4 = 4 \left( 4 q ^ { 2 } + 4 q + 1 \right) = 4 m{/tex}
{tex}( a ) ^ { 2 } = ( 4 q + 3 ) ^ { 2 } = 16 q ^ { 2 } + 24 q + 9 = 4 \left( 4 q ^ { 2 } + 6 q + 2 \right) + 1 = 4 m + 1{/tex}
{tex}\therefore {/tex} the square of any +ve integer is of the form 4q or 4q + 1
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