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**tony123****Member**- Registered: 2007-08-03
- Posts: 218

solve in integers

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

(a-b-1)^3+(b-c-2)^3+(c-a+3)^3=[(a)-(b+1)]^3+[(b+1)-(c+3)]^3+[(c+3)-(a)]^3. Let (b+1)=d & (c+3)=e. Then, (a-d)^3+(d-e)^3+(e-a)^3=18. Let a-d=x & d-e=y. Then, a=x+d & e-a=e-x-d=-x-d+e=-x-(d-e)=-x-y=-(x+y). So, x^3+y^3-(x+y)^3=18 or -3xy(x+y)=18 or xy(x+y)=-6. Since all numbers are integers, ((xy),(x+y))=(1,-6), (-1,6),(6,-1),(-6,1) respectively and the only acceptable combination is (-6,1) resulting to (x,y)=(3,2) or (-2,3). From the first pair we get d-e=-1 & d-e=-2, so we select the second pair, thus x=-2 & y=3, so a-d=2, d-e=3, e-a=1 or b=a+1 & c=a-4. So, (a,b,c)=(a,a+1,a-4). There are infinite solutions, i.e. (a,b,c)=(2,3,-2),(3,4,-1),(50,51,46) etc.

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

Sorry about the smilies , I forgot to turn the thing off. So, here is the proper solytion; (a-b-1)^3+(b-c-2)^3+(c-a+3)^3=[(a)-(b+1)]^3+[(b+1)-(c+3)]^3+[(c+3)-(a)]^3. Let (b+1)=d & (c+3)=e. Then, (a-d)^3+(d-e)^3+(e-a)^3=18. Let a-d=x & d-e=y. Then, a=x+d & e-a=e-x-d=-x-d+e=-x-(d-e)=-x-y=-(x+y). So, x^3+y^3-(x+y)^3=18 or -3xy(x+y)=18 or xy(x+y)=-6. Since all numbers are integers, ((xy),(x+y)) = (1,-6), (-1,6),(6,-1),(-6,1) respectively and the only acceptable combination is (-6,1) resulting to (x,y) = (3,2) or (-2,3). From the first pair we get d-e=-1 & d-e=-2, so we select the second pair, thus x=-2 & y=3, so a-d=2, d-e=3, e-a=1 or b=a+1 & c=a-4. So, (a,b,c) = (a,a+1,a-4). There are infinite solutions, i.e. (a,b,c) = (2,3,-2),(3,4,-1),(50,51,46) etc.

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**bitus****Member**- Registered: 2008-10-12
- Posts: 20

Yesterday's solution was incomplete (due to some excellent wine...) and, in fact, wrong. Here's the full stuff; First, (x,y) = (3,2) is valid and we get (a,b,c) = (a,a-4,a-5). BUT, there are four more combinations for ((xy),(x+y)) ; (2,-3) , (-2,3) , (3,-2) , (-3,2). Since we are looking for integers, we heve four more pairs for (x,y) ; (-1,3) , (3,-1) , (-2,-1) , (-1,-2) or (a,b,c) = (a,a+1,a) , (a,a,a-5) , (a,a,a) , (a,a-4,a-4). In fact, there are six groups of possible solutions, because six is the number of possible combinations for the three parenthesis (3!) to be 3,-1,-2 since (3)^3+(-1)^3+(-2)^3 = 18. Obviously, you can solve the problem starting from this last remark.

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