Square And Cubes Challenging You
Find the sum of all possible solutions of a 2 − b 2 for the following conditions:
- a and b are positive integers less than 1,000,
- a 3 − b 3 is perfect square number,
- a 3 − b 3 − 3 a b = 1 .
Example : If possible solutions of ( a , b ) are ( 2 , 1 ) and ( 3 , 2 ) then the sum of all ( a 2 − b 2 ) = ( 4 − 1 ) + ( 9 − 4 ) = 8 .
The answer is 224.
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1 solution
It is not clear that a − b = 1 . Letting a − b = x , we have x 3 + 3 a b x = 1 + 3 a b , which gives us 0 = x 3 − 1 + 3 a b x − 3 a b = ( x − 1 ) ( x 2 + x + 1 + 3 a b ) . In order to conclude that x − 1 = 0 , you will need to show that x 2 + x + 1 + 3 a b = 0 . However, this could happen when a = − 1 , b = 1 .
Also, how did you conclude that the only way for 3 b 2 + 3 b + 1 to be a perfect square is for b = 7 , 1 0 4 ?
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As per the conditions,a and b are positive integers .
Hence in the expression x 2 + 1 + x + 3 a b , x 2 + 3 a b + 1 > x => x 2 + 1 + x + 3 a b > 0
So,a-b=1
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Yes, that's the way to explain why "it is clear that a − b = 1 ". Can you include this into your solution?
I see that you added the condition "less than 1000". As mentioned, you are solving the Pell's equation ( 2 c ) 2 − 3 ( 2 b + 1 ) 2 = 1 . The Pell's equation X 2 − 3 Y 2 = 1 has initial solution ( 2 , 1 ) ), which generates ( 2 , 1 ) , ( 7 , 4 ) , ( 2 6 , 1 5 ) , ( 9 7 , 5 6 ) , ( 3 6 2 , 2 0 9 ) , ( 1 3 5 1 , 7 8 0 ) , ( 5 0 4 2 , 2 9 1 1 ) … . Hence, this corresponds to 2 b + 1 = 1 , 1 5 , 2 0 9 , in which case we have b = 7 , 1 0 4 as the only solutions under 1000.
The Pell's equation also gives us infinitely many solutions. The next one would be b = 1 4 5 5 , and they grow exponentially.
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@Calvin Lin – I am not able to edit the solution and it seems only one solution can be added by one person.How can I make the changes ???
Thank you for the solution regarding "Pell's equation" .I need to study the Pell's equation .
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@Vinay Sipani – You can edit your solution (and comments), by clicking on the pencil icon that is to the top right corner, next to the "posted time".
Sorry...I am not sure whether there exist solutions other than 7 and 104.
I have made the changes in the problem...
Thank You.
@Calvin Lin Does there exist more than 2 solutions...Please give a solution to this which is related to this problem LINK TO NOTE
Sir, I am not able to raise objection. I see no way to do it since there is no record button. (1, 0) is also a solution according to me. 0 is a positive integer!!!!!
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I don't think 0 is a positive integer. So far what I have seen is if 0 is included they write "Non-negative" integers.
a, b >0 they are both positive integers
Given: a 3 − b 3 − 3 a b = 1
= > ( a − b ) 3 + 3 a b ( a − b ) = 1 + 3 a b Letting ( a − b ) = x ,we have
x 3 + 3 a b x = 1 + 3 a b = > ( x − 1 ) ( x 2 + x + 1 ) + 3 a b ( x − 1 ) = 1 = > ( x − 1 ) ( x 2 + x + 1 + 3 a b ) = 0 Thus,either x − 1 = 0 or x 2 + x + 1 + 3 a b = 0
Consider x 2 + x + 1 + 3 a b , From the conditions,a and b are positive integers.
= > x 2 > x ; 3 a b > 0
= > x 2 + x + 1 + 3 a b = 0
Therefore,x=(a-b)=1
Hence, ( a 3 − b 3 ) = a 2 + b 2 + a b = ( b + 1 ) 2 + b 2 + ( b + 1 ) b = 3 b 2 + 3 b + 1
Let 3 b 2 + 3 b + 1 = x 2 = > 3 b 2 + 3 b + ( 1 − x 2 ) = 0
Discriminant must be a perfect square. = > 3 2 − 4 × 3 × ( 1 − x 2 ) = y 2 = > y 2 − 1 2 x 2 + 3 = 0
This is the Pell's equation which gives the solutions (x,y)=(13,45) and (181,627) less than 1000.
=>Possible values are ( a , b ) = ( 8 , 7 ) a n d ( 1 0 5 , 1 0 4 ) = > Sum of all a 2 − b 2 = ( 8 + 7 ) + ( 1 0 5 + 1 0 4 ) = 2 2 4