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When n is an positive integer , 2n is an even number , and the next even number is 2n+2,

Hence explain why the squares of the 2 consequtive even numbers is always a multiple of 4,

Can I get a hint on the method how to solve this ? Thanks !

user307640
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    $(2m)^2= 4m^2$... – lhf Jul 21 '16 at 02:45
  • Observe that once you've proved it for $2n$, then you've proved it for all even numbers, so it's redundant to show it for $2n+2$. The property has absolutely nothing to do with $2n$ and $2n+2$ being consecutive. – Théophile Jul 21 '16 at 04:04

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Squaring $2n$ we have $(2n)^2=4n^2$, which is clearly divisible by $4$. Squaring $2n+2$ we have $(2n+2)^2=4n^2+8n+4=4(n^2+2n+1)$, which is also clearly divisible by $4$.

Dave
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