# Diophantine equations - Perfect square and Perfect cube related

Solve following Diophantine equations:

$1) \ a^3-a^2+8=b^2$

2) $a, \ b,\ c \in \mathbb{Z^+}$$\frac{a^3}{(b+3)(c+3)} + \frac{b^3}{(c+3)(a+3)} + \frac{c^3}{(a+3)(b+3)} = 7$$ 3)$a^3-8=b^2$In Problem 2 I tried to use inequality, then I can 'limit' that:$25 \ge a+b+c$and$a^3 + b^3 + c^3 \ge 112$Please use elementary way to solve it, I haven't studied elliptic curve yet, thanks. - As a hint for Problem 1), try adding 4 to both sides of the equation and see what happens... – Mike Bennett Feb 12 '13 at 19:40 Sorry professor but could you tell me more detail,I am not good in diophantine equations ? I have tried add 4 to both sides before but it led to$(x+2)(x^2 - 3x + 6)=y^2 + 4$I also have$12 | y, x \equiv 2 \ (mod\ 3)$but seem didn't help. – Xeing Feb 13 '13 at 9:44 Next step : which primes can divide$y^2+4$? – Mike Bennett Feb 13 '13 at 20:15 ## 1 Answer (2) Extremely ugly solution. You have $$a^3(a+3)+b^3(b+3)+c^3(c+3)=7(a+3)(b+3)(c+3)$$ or $$(x-3)^3x+(y-3)^3y+(z-3)^3z=7xyz$$ Since it is symmetric, we can look for the solutions where$x \geq y \geq z$. It is easy to check that for$x>15$we have$(x-3)^3>7x^2$. This shows that$4 \leq x \leq 14$. For each particular$x$you get a simpler solution which can be solved the same way. P.S. It probably also helps observing that modulo 3 you have $$(w-3)^3w \equiv 0,1 \pmod 3$$ Then, if none of$x,y,z$is divisible by$3$the LHS is 0$\pmod 3$which is not possible. If one of$x,y,z$is divisible by$3\$, then all of them must be divisible by 3, and looking at the equation it follows that one of them is divisible by 9.

This should solve the equation.

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