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# Integral solutions

Here are a few interesting problems :

$$(1)$$ Prove that there are infinitely many triples $$(x,y,z) \in \mathbb{Z^{+}}$$ such that

$x^3+y^4=z^7$

$$(2)$$ Show that there are no solutions $$(a,b,c) \in \mathbb{Z}$$ such that

$a^2+b^2-8c=6$

$$(3)$$ Determine all triples $$(x,y,z) \in \mathbb{Z^{+}}$$ satisfying

$2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4 = 576$

Note by Ankit Kumar Jain
4 months ago

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$$(1)$$ :

Set $$x=2^{4a},y=2^{3a},z=2^{b}$$ , which leads on simplification to the linear diophantine equation $$7b-12a=1$$ which has infinitely many solutions since $$(12,7)=1$$ . One such being $$(x,y,z)=(2^{16},2^{12},2^{7})$$

$$(2)$$ :

Given, $$\displaystyle a^2+b^2=8c+6\;\equiv\; 6\;(mod \; 8)$$ . Now it is evident that for any integer $$n$$ the relation,

$$\displaystyle n^2\;\equiv\;0,1,4\;(mod\; 8)$$ is satisfied which clearly contradicts $$a^2+b^2\;\equiv\;6\;(mod \;8)$$ . Thus no solutions. · 4 months ago

Good ones. · 4 months ago

You can try out my note Real Roots. · 4 months ago

$$(3) :$$

\begin{align}(x+y+z)(x+y-z)(y+z-x)(z+x-y)&=2x^2y^2+2y^2z^2+2z^2x^2-x^4-y^4-z^4\\ \implies (x+y+z)(x+y-z)(y+z-x)(z+x-y)&=576\\\\ \end{align}

$$\text{Let, } x+y+z=a\\$$

\begin{align}a(a-2x)(a-2y)(a-2z)&=576 \hspace{7mm}\color{blue}\small (1)\\\\ \implies \text{All the factors are }\text{even}&\hspace{7mm}\color{blue}\small (2)\hspace{7mm}\text{Since a,a-2x,a-2y,a-2z are all of the same parity}\\\\ a-2x+a-2y+a-2z&=3a-2(x+y+z)\\\\ &=3a-2a=a\end{align}\\\\ \implies \text{The largest factor is the sum of the other 3 factors}\hspace{7mm}\color{blue}\small (3)

\begin{align}\text{The only unordered pair that satisfy}\small \color{blue}(1),(2) \normalsize\color{black}\text{ and }\small\color{blue}(3) \color{black}\normalsize \text{is,}(12,6,4,2)\\ \implies (x,y,z) \text{can take the values } (3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4),(5,4,3)\end{align} · 4 months ago

Nice solution. · 4 months ago