Maybe Ramanujan; Maybe not!
a 3 + b 3 = c 3 + d 3 For the minimum positive value of expression a 3 + b 3 find the value of a + b + c + d where a , b , c , d are distinct integers.
The answer is 8.
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2 solutions
Moderator note:
When checking cases with a computer, we should explain why it is sufficient to look at the finite range of values that a computer can check.
Can you explain why that is the minimium value?
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Sir I came across it somewhere on the Internet and then used a java programme to reconfirm it ,can you please provide a solution to it . I can post the java programme if you insist.
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Go ahead and post the java program. Some extent of trial and error will be involved, and we can demonstrate that checking is required.
Nice question BTW
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@Calvin Lin – Thank you😊 I will add it up to my solution.
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@Chaitnya Shrivastava – @Calvin Lin Done sir but is there any proper number theory solution to it?
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@Chaitnya Shrivastava – I do not know. There could be.
Note that in your java code, you should justify why you only need to look at numbers in that range. In fact, we only need to search for single digit integers (and their negative values).
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@Calvin Lin – I used those values as they were the maximum and minimum value of long data type but I understood what you mean.
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@Chaitnya Shrivastava – So, because we know that 91 is already an answer, we just want to find loose conditions for when 0 < a 3 + b 3 ≤ 9 0
For example, one possibility is that a 3 + b 3 = ( a + b ) ( a 2 − a b + b 2 ) , and since the second term is a positive integer, we can conclude that a + b ≤ 9 0 . Also, since a 3 + b 2 ≥ ∣ a ∣ 3 − ( ∣ a ∣ − 1 ) 3 , this gives us another (really restrictive) condition to look at.
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@Calvin Lin – So we can arrive at the result by checking values in these conditions and why should a 2 + b 2 − a b necessarily be positive?
I could not findout the meaning of the question! can a=0, b=1,both are integer. a^3 + b^3=1 is minimum positive value! And a+b+c+d=0+1+0+1=2!
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Here the condition says a 3 + b 3 = c 3 + d 3 so you have to find a minimum value that satisfies both conditions with of course a not equal to c & b not equal to d or vice versa. I have added the details.
I also find a solution that a+b+c+d=8 ,although my a^3+b^3 is not the minimum positive value. That is 9^3 + (-8)^3 = 6^3 + 1^3 = 217 ,and 9 + (-8) + 6 + 1 =8
so, this problem have 2 ways to get the answer 8.
When we cosider integers there are a finite number of positive solutions less than 1729 and infinite number of them if we consider negative value but if you look for the minimum positive value its 91 as I have mentioned in my solution. You were lucky though😅
Here is a third way: ( − 1 5 ) 3 + 1 6 3 = ( − 2 ) 3 + 9 3 = 7 2 1 and (-15)+16+(-2)+9=8
Firstly a 3 + b 3 does not equal to 1729 because 1729 is the minimum value when a,b,c&d are natural numbers i.e. 1 2 3 + 1 3 = 1 0 3 + 9 3 = 1 7 2 9 BUT HERE WE ARE DEALING WITH INTEGERS Therefore 6 3 + ( − 5 ) 3 = 4 3 + 3 3 = 9 1 So the minimum value of a 3 + b 3 is 91 and a + b + c + d = 8 we can confirm the value using a programme