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Question :

Show that $a^3 + b^4 + c^5 = d^7$ has infinitely many solutions .


My try :

I knew that the number 3 has this special property $$3^n +3^n +3^n = 3^{n+1}$$

With this i proved that it has infinitely many solution.Now my question is, is there any way to prove this without using this special property. Please help me.

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Hint If $(a, b, c, d)$ is a solution, substitution shows that so are the infinitely many tuples $$(\lambda^{4 \cdot 5 \cdot 7} a, \lambda^{3 \cdot 5 \cdot 7} b, \lambda^{3 \cdot 4 \cdot 7} c, \lambda^{3 \cdot 4 \cdot 5} d), \qquad \lambda \in \Bbb Z .$$ So, it suffices to find a single nonzero solution,

It's easy to find a nonzero solution (e.g., $(1,0,0,1)$), and it's not too much harder to find a solution in positive integers: Using the decomposition $2^{m + 2} = 2^{m + 1} + 2^{m + 1} = 2^{m + 1} + 2^m + 2^m$ (this observation is used in achille hui's solution, too), substituting shows that $$(2^{\alpha}, 2^{\beta}, 2^{\gamma}, 2^{\delta})$$ is a solution if, for example, $$3 \alpha = 5 \gamma = m, \qquad 4 \beta = m + 1, \qquad 7 \delta = m + 2$$ for some integer $m$. The first condition implies $15 \mid m$, and then the second condition gives $m = 15(4 k + 1)$ for some integer $k$. The smallest positive integer solution that also satisfies the last condition is then $k = 1$, or $m = 75$. Substituting gives that the corresponding solution is $$(2^{25}, 2^{19}, 2^{15}, 2^{11}) = (33\,554\,432, 524\,288, 32\,768, 2\,048) .$$

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  • $\begingroup$ Really good analysis, thanks $\endgroup$ – Oldboy Oct 31 '18 at 15:58
  • $\begingroup$ Cheers!$\!\!\!$ $\endgroup$ – Travis Willse Oct 31 '18 at 16:41
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I think that your solution is the simples one. It can be easily proved that:

$$a=3^{20k}, \space b=3^{15k},\space c=3^{12k},\space d=3^{60k+1 \over 7}$$

satisfy the equation for infinitely many values of $k$ such that $k\equiv 5 \bmod 7$.

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Here is another family of solutions that doesn't involve the number $3$: $$(a,b,c,d) = ( 2^{140k + 88}, 2^{105k + 66}, 2^{84k + 53}, 2^{60k+38} )$$

The basic idea is search for $a,b,c,d$ among powers of two. It is easy to see if they satisfy $$a^3 = b^4 = \frac12 c^5 = \frac14 d^7$$ then the equality $a^3 + b^4 + c^5 = d^7$ follows.

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