259 followers problem

Algebra Level 5

$\begin{cases} a^3 + 12ab+ 64=b^3 \\ a^4 + 256=b^4 \end{cases}$

If $(a_1,b_1),(a_2,b_2) ,\ldots , (a_n,b_n)$ are real numbers which satisfy the system of equations above, find the value of

$\displaystyle 1 + 27 \sum_{j=1}^n (b_j^3 - a_j^3).$

The answer is 1729.

2 solutions

Dev Sharma
Nov 10, 2015

$a^3 - b^3 + 64 + 12ab = 0$

$\dfrac{1}{2}(a - b + 4)((a+b)^2 + (b + 4)^2 + (4-a)^2) = 0$

So there are two possible cases

CASE1 When $(a+b)^2 + (b + 4)^2 + (4-a)^2 = 0$

Then $a = 4, b = -4$

But it doesnt satisfy

CASE2 When $a - b + 4 = 0$ that is $a+4=b$

Putting this value in the second equation :

$a^4 + 256 = (a+4)^4$

Solving it, we get a cubic equation, we can found the only possible pair is $(a,b)=(0,4)$

So our answer is 1729

I too did it exactly the same way!!

Surya Prakash - 5 years, 7 months ago

Aditya Kumar
Nov 12, 2015

The 2nd equation can be thought of as the fermat last theorem so it must not have any positive integral solutions which gave us solution as a=0 and b=4. But the question has asked in the domain of all real numbers but i dont know how to prove that any non integral solution is not possible.

259 followers problem

The answer is 1729.

2 solutions

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