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Well, I make some Diophantine equations sometimes. I like to see different solutions to them. Now suppose that in the following equation, all variables are some digits between 0 and 9. Find all solutions.$$\overline{xy.z}^2=800 \ \overline{r.styz}$$I have a solution to this problem and I would love to see your way of looking at this.

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  • $\begingroup$ The letters are digits? What do the bars mean? $\endgroup$
    – mvw
    Commented Mar 20, 2017 at 9:44
  • $\begingroup$ Is it $800 \times \overline{r.styz}$? $\endgroup$
    – S.C.B.
    Commented Mar 20, 2017 at 9:45
  • $\begingroup$ @mvw It means that for example $\overline{xy.z}$ is a number like $25.6$. $\endgroup$
    – Ghartal
    Commented Mar 20, 2017 at 9:48
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    $\begingroup$ solve $m^2=8n $ for $ 0\leq m\leq 999 $ and $ 0\leq n\leq 99999$. $\endgroup$
    – Lozenges
    Commented Mar 20, 2017 at 10:22
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    $\begingroup$ Wait :)) $m$ and $n$ are dependent! How can you consider this dependency? $\endgroup$
    – Ghartal
    Commented Mar 20, 2017 at 11:06

2 Answers 2

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Let $m=10(xy.z)$ and $n=10000(r.styz)$ We have to solve the system $$m^2=8n$$ $$m\equiv n(100)$$

where $0\leq m\leq 999$ and $0\leq n\leq 99999$

Now, $n=m+100k$ for some $k$ gives $m^2=8m+800k$. Therefore we have to solve

$$m^2\equiv 8m(800)$$

Conversely, any solution of this congruence will give a solution of the above system if we let $n=\frac{m^2}{8}$ provided $0\leq m\leq 894$ to make sure $0\leq n\leq 99999$

Next, show $m$ is divisible by $8$. We know $m$ is divisible by $4$ so let $m=4t$. We get $16t^2-32t$ divisible by $800$ therefore divisible by $32$ and so $t$ is divisible by $2$ and $m$ is divisible by $8$

Finally, let $m=8u$ , $0\leq u\leq 111$ and show that this leads to $u^2\equiv u (25)$ which means either $u$ or $u-1$ must be a multiple of $25$ since they can't both be divisible by $5$

$$u=\{0,25,50,75,100,1,26,51,76,101\}$$

$$m=\{0,200,400,600,800,8,208,408,608,808\}$$

$$n=\{0,5000,20000,45000,80000,8,5408,20808,46208,81608\}$$

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  • $\begingroup$ Bravo. Beautiful solution :) $\endgroup$
    – Ghartal
    Commented Mar 20, 2017 at 17:32
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$$\left(10x+y+\frac{z}{10}\right)^2=800\left(r+\frac{s}{10}+\frac{t}{100}+\frac{y}{1000}+\frac{z}{10000}\right)$$ is equivalent to $$100x^2+y^2+\frac{z^2}{100}+20xy+2xz+\frac{yz}{5}=800r+80s+8t+\frac 45y+\frac{2}{25}z$$ Multiplying the both sides by $100$ gives $$10000x^2+100y^2+z^2+2000xy+200xz+20yz=80000r+8000s+800t+80y+8z$$ Since $z$ has to be even, letting $z=2z'$ where $z'=0,1,2,3,4$ and dividing by $4$ give $$2500x^2+25y^2+z'^2+500xy+100xz'+10yz'=20000r+2000s+200t+20y+4z'\tag1$$ from which we have $$z'^2-4z'\equiv 0\pmod 5\implies z'=0,4$$

Case 1 : If $z'=0$, then $$(1)\iff 500x^2+5y^2+100xy=4000r+400s+40t+4y\tag2$$from which we have $$4y\equiv 0\pmod 5\implies y=0,5$$

Case 1-1 : If $y=0$, then $$(2)\iff 25x^2=200r+20s+2t$$ Setting $x=2x'$ where $x'=0,1,2,3,4$ and dividing by $2$ give $$50x'^2=100r+10s+t$$ Since $t$ has to be $0$, we have $$5x'^2=10r+s$$ giving $s=0,5$, so $$(s,r,x')=(0,0,0),(0,2,2),(0,8,4),(5,0,1),(5,4,3)$$

Case 1-2 : If $y=5$, then $$(2)\iff 100x^2+25+100x=800r+80s+8t+4$$ The LHS is odd while the RHS is even, which is impossible.

Case 2 : If $z'=4$, then $$(1)\iff 500x^2+5y^2+100xy+80x=4000r+400s+40t-4y\tag3$$ from which $$-4y\equiv 0\pmod 5\implies y=0,5$$

Case 2-1 : If $y=0$, then $$(3)\iff 25x^2+4x=200r+20s+2t$$ Setting $x=2x'$ where $x'=0,1,2,3,4$ gives $$50x'^2+4x'=100r+10s+t$$from which we have $$4x'-t\equiv 0\pmod 5\quad\text{and}\quad \frac{4x'-t}{5}\equiv 0\pmod 2$$$$\implies (x',t)=(0,0),(1,4),(2,8),(3,2),(4,6)$$ giving $$(x',t,r,s)=(0,0,0,0),(1,4,0,5),(2,8,2,0),(3,2,4,6),(4,6,8,1)$$

Case 2-2 : If $y=5$, then $$(3)\iff 100x^2+25+100x+16x=800r+80s+8t-4$$ The LHS is odd while the RHS is even, which is impossible.

Therefore, the answer is $$\color{red}{(r,s,t,x,y,z)=(0,0,0,0,0,0),(2,0,0,4,0,0),(8,0,0,8,0,0),(0,5,0,2,0,0),(4,5,0,6,0,0),}$$$$\color{red}{(0,0,0,0,0,8),(0,5,4,2,0,8),(2,0,8,4,0,8),(4,6,2,6,0,8),(8,1,6,8,0,8)}$$

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  • $\begingroup$ Quite similar to what I've done! Thanks :) $\endgroup$
    – Ghartal
    Commented Mar 20, 2017 at 11:08

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