Algebra with indices
- Thread starter Seabo04
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HallsofIvy
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Have you looked at different values of n?
If n= 1, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^3= 2+ 8= 10= 9+ 1= 3^2+ 1^2\)
If n= 2, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^5= 2+ 32= 34= 25+ 9= 5^2+ 3^2\).
If n= 3, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^7= 2+ 128= 130= 81+ 49= 9^2+7^2\).
If n= 4, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^9= 2+ 512= 514= 289+ 225= 17^2+ 15^2\).
So we have squares of 3, 5, 9, and 17 added to squares of 1, 3, 7, and 15. Do you see any patterns? I notice that 5= 2(3)- 1, 9= 2(5)- 1, and 17= 2(9)- 1 and that 3= 2(1)+ 1, 7= 2(3)+ 1, and 15= 2(7)+ 1.
If n= 1, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^3= 2+ 8= 10= 9+ 1= 3^2+ 1^2\)
If n= 2, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^5= 2+ 32= 34= 25+ 9= 5^2+ 3^2\).
If n= 3, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^7= 2+ 128= 130= 81+ 49= 9^2+7^2\).
If n= 4, \(\displaystyle 2+ 2^{2n+1}= 2+ 2^9= 2+ 512= 514= 289+ 225= 17^2+ 15^2\).
So we have squares of 3, 5, 9, and 17 added to squares of 1, 3, 7, and 15. Do you see any patterns? I notice that 5= 2(3)- 1, 9= 2(5)- 1, and 17= 2(9)- 1 and that 3= 2(1)+ 1, 7= 2(3)+ 1, and 15= 2(7)+ 1.