Sizing up factors
Is it possible to write 3 2 0 1 6 + 4 2 0 1 7 as the product of two integers, both of which are over 2 0 1 8 1 8 3 ?
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2 solutions
@Sharky Kesa Same way.....but shouldn't the last word in your solution be proved.......?
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I don't think so. The entire proof was in present tense, so the final statement should be in present tense.
Essentially, we want to prove that 3 2 0 1 6 + 4 2 0 1 7 > ( 2 0 1 8 1 8 3 ) 2 = 2 0 1 8 3 6 6 .
I will prove this by reverse engineering this inequality. We start with: 4 0 3 4 2 4 0 3 4 4 2 0 1 7 4 2 0 1 7 + 3 2 0 1 6 > > > > 4 0 2 6 2 4 0 2 6 = ( 2 1 1 ) 3 6 6 2 0 4 8 3 6 6 > 2 0 1 8 3 6 6 2 0 1 8 3 6 6
This doesn't prove that there exist factors both of which are over 2 0 1 8 1 8 3 .
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You said product of "2 numbers", not product of "2 integers".
Note that a 4 + 4 b 4 = ( a 2 + 2 b 2 − 2 a b ) ( a 2 + 2 b 2 + 2 a b ) by the Sophie-Germain Identity. Further note that a 2 + 2 b 2 − 2 a b = ( a − b ) 2 + b 2 > b 2 and a 2 + 2 b 2 + 2 a b = ( a + b ) 2 + b 2 > b 2 . Thus, 3 2 0 1 6 + 4 2 0 1 7 = 3 2 0 1 6 + 4 ⋅ 4 2 0 1 6 = ( 3 1 0 0 8 + 2 ⋅ 4 1 0 0 8 + 2 ⋅ 3 5 0 4 ⋅ 4 5 0 4 ) ( 3 1 0 0 8 + 2 ⋅ 4 1 0 0 8 − 2 ⋅ 3 5 0 4 ⋅ 4 5 0 4 ) .
It can be seen that 3 1 0 0 8 + 2 ⋅ 4 1 0 0 8 ± 2 ⋅ 3 5 0 4 ⋅ 4 5 0 4 > 4 1 0 0 8 = 2 2 0 1 6 > ( 2 1 1 ) 1 8 3 = 2 0 4 8 1 8 3 > 2 0 1 8 1 8 3 , so both numbers satisfy the bound. Thus, proven.