2009 JBMO Shortlist #A1

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Determine all integers $a, b, c$ satisfying the identities $$a + b + c = 15$$ 

$$(a - 3)^3 + (b - 5)^3 + (c -7)^3 = 540.$$



Note that if $x, y, z$ are integers such that $x+y+z=0$, then $$x^3+y^3+z^3=x^2+y^3+(-x-y)^3=x^3+y^3+x^3+y^3-3xy(x+y)=3xyz.$$

Also, since $a+b+c=15$, we have $(a-3)+(b-5)+(c-7)=0$. Using the fact we noted above, we see that $$540\underset{\text{given}}{=}(a-3)^3+(b-5)^3+(c-7)^3=3(a-3)(b-5)(c-7)$$

$$\implies (a-3)(b-5)(c-7)=180=2^3\cdot 3^2\cdot 5.$$

But $(a-3)+(b-5)+(c-7)=0$, so we must have $(a-3)(b-5)(c-7)=4\cdot 5\cdot 9$. 

We have four possible cases from here, namely 

$$(a-3, b-5, c-7)=(-4,-5,9)$$ 

$$(a-3, b-5, c-7)=(-5,-4,9)$$

$$(a-3, b-5, c-7)=(4,5,9)$$

$$(a-3, b-5, c-7)=(5,4,9)$$

Solving these $4$ system of equations gives us $$\boxed{(a,b,c)=(-1,0,16), (-2,1,16), (7, 10, -2), (8, 9, -2)}.$$

$\square$

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