Five variables, one solution
It is known that there is such a number s such that if real numbers a , b , c , d are all neither 0 nor 1, satisfying a 1 + b 1 + c 1 + d 1 = s and a + b + c + d = s , then 1 − a 1 + 1 − b 1 + 1 − c 1 + 1 − d 1 = s Find s .
The answer is 2.
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1 solution
Direct calculation gives (s-2)[s-2+sabcd-(ab+ac+ad+bc+bd+cd)]=0, which implies s=2 for any a, b, c, d satisfying the given conditions. Good work. There is a little typo though. It will be s-2+sv-u instead of s-2+u-sv
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Typo fixed. Certainly direct calculation will show that identity, but it is not at all obvious why you should try for that identity in particular. My approach gives a reason for the formula!
Very nice, as always. I did come across one shortcut, in case you're interested (I don't think it's worth writing up as a separate solution, though).
We can write
1 − a 1 + 1 − b 1 + 1 − c 1 + 1 − d 1 = ( 1 − a ) ( 1 − b ) ( 1 − c ) ( 1 − d ) ( 1 − b ) ( 1 − c ) ( 1 − d ) + ( 1 − a ) ( 1 − c ) ( 1 − d ) + ( 1 − a ) ( 1 − b ) ( 1 − d ) + ( 1 − a ) ( 1 − b ) ( 1 − c ) = f ( 1 ) f ′ ( 1 )
which immediately gives 4 − 3 s + 2 u − s v = s ( 1 − s + u − s v + v ) without having to introduce the function g .
It appears that many extremely good approximations can be made, that don't result in s = 2 . For example:
a ≈ − 0 . 3 9 2 9 2 3 b ≈ 0 . 3 6 5 3 5 5 c ≈ 0 . 3 5 4 1 2 9 d ≈ 3 . 0 2 0 3 8 0 s ≈ 3 . 3 4 6 9 4 2
What shall we make of this? I interpreted the question as meaning "Find ( a , b , c , d , s ) such that the following three equations are simultaneously satisfied"
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This is a misinterpretation of the question. In general, for some value of s , there are many possible sets of a , b , c , d such that a + b + c + d = s and a − 1 + b − 1 + c − 1 + d − 1 = s . These various sets given many possible values for ( 1 − a ) − 1 + ( 1 − b ) − 1 + ( 1 − c ) − 1 + ( 1 − d ) − 1 . There is only one value of s which, no matter what the values of a , b , c , d , gives ( 1 − a ) − 1 + ( 1 − b ) − 1 + ( 1 − c ) − 1 + ( 1 − d ) − 1 = s , namely s = 2 .
If a + b + c + d = a − 1 + b − 1 + c − 1 + d − 1 = s , then a , b , c , d are the roots of the quartic equation f ( X ) = X 4 − s X 3 + u X 2 − s v X + v = 0 for some u , v . Then ( 1 − a ) − 1 , ( 1 − b ) − 1 , ( 1 − c ) − 1 , ( 1 − d ) − 1 will be the roots of the equation f ( 1 − Y − 1 ) = 0 , and so the roots of g ( Y ) = Y 4 f ( 1 − Y − 1 ) = ( Y − 1 ) 4 − s Y ( Y − 1 ) 3 + u Y 2 ( Y − 1 ) 2 − s v Y 3 ( Y − 1 ) + v Y 4 = 0 Now g ( Y ) is a quartic polynomial, and its first two terms are g ( Y ) = ( 1 − s + u + ( 1 − s ) v ) Y 4 − ( 4 − 3 s + 2 u − s v ) Y 3 + ⋯ and since the sum of the roots of this quartic is also s , we deduce that 4 − 3 s + 2 u − s v s 2 − 4 s + 4 + ( 2 − s ) u + s ( s − 2 ) v ( s − 2 ) [ s − 2 − u + s v ] = s [ 1 − s + u + ( 1 − s ) v ] = 0 = 0 and this equation will be true, whatever the values of u and v , provided that s = 2 .