mirror equations
a 3 + b 2 + c = 26
c 3 + b 2 + a =746
the set i am looking for forms the first three elements or terms of a very identifiable series
find the sum of a,b and c where a,b,c are positive integers
The answer is 14.
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2 solutions
since the first equation has a cube and is also numerically equivalent to 26, only two possibilities for a are 2,1. Moreover, if a = 2, we get c 3 + b 2 = 7 4 4 which has no integral solutions due to obvious reasons. thus a=1 now c 3 + b 2 = 7 4 5 and c + b 2 = 2 5
subtracting the two equations so obtained, we get c ( c + 1 ) ( c − 1 ) = 7 2 0 which can be written as 9 ( 1 0 ) ( 8 )
thus,c is 9 and now b is obtained as 4
thus the solution set is the subset of the set of square numbers (1,4,9)
thus a + b + c = 1 4
It didn't mention that they must be integers. Nor did they mention that they must be positive.
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i am sorry . i forgot to mention i will mention it immediately
What are the obvious reasons that c 3 + b 2 = 7 4 4 has no solutions?
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to arrive the the equation we have taken two cases where a is 1 and 2.
on taking a as 2 we get the equation u want to know about
also a value of two for a gives us b 2 + c = 1 6 this leaves us with just three possibilities for b namely 1,2,3
substituting these values for b we do not get integral solutions for c
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You should seek to provide complete explanations of your statements. As opposed to
If a = 2 then c 3 + b 2 = 7 4 4 has no integer solutions for obvious reasons.
It would be better to say
If a = 2 then c 3 + b 2 = 7 4 4 has no positive integer solutions as we check through cases of c = 1 , … 9 .
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@Calvin Lin – Thank you I will keep that in mind from now on
It is evident in the equations that b cannot be larger than 5 , even more obvious for a , thus making it obvious that c should be as large as possible to compensate for equation 2. The nearest cube to 7 4 6 is 7 2 9 , so c = 9 , and that being said, it follows that a = 1 and b = 4 , because no other combination fits.
This makes a + b + c = 1 4