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Greg
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Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
greg1313 said:Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
mente oscura said:Hello.
Regards.
MarkFL said:It is fine to use a link to back up or justify a technique, but can you show how would apply that to solve the problem?
greg1313 said:Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
I'm trying a brute force approach and I've got that a = 3n, for n = 1, 2, 3...greg1313 said:Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
If $x\geqslant y\geqslant0$ then clearly $\lvert x - y \rvert \leqslant x$. But I don't see why $\lvert x - y \rvert \leqslant y$. For example, if $y=1$ and $x=3$ then $|x-y| = 2$, which is greater than $y$. Am I missing something here? [/sp]Bacterius said:Now for the key step: $(x - y)^2$ cannot be any larger than $xy$, because $\lvert x - y \rvert \leq x$ and $\lvert x - y \rvert \leq y$
Opalg said:There is one step in Bacterius's solution that I don't follow.
[sp]
If $x\geqslant y\geqslant0$ then clearly $\lvert x - y \rvert \leqslant x$. But I don't see why $\lvert x - y \rvert \leqslant y$. For example, if $y=1$ and $x=3$ then $|x-y| = 2$, which is greater than $y$. Am I missing something here? [/sp]
greg1313 said:Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
How do you know what the series must be if you don't already know k? For instance, how do you know k = 5 doesn't work?Albert said:now we create two sequences $a_n,b_n\in N$
$x^2-kxy+y^2+1=0----(2)$
the solution of (2) will be:
$x=a_1=1,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y=b_1=2$
$x=a_2=2=b_1,y=b_2=5$
$x=a_3=5=b_2,y=b_3=13$
gven :$x,y\in N$topsquark said:I'm confused.
How do you know what the series must be if you don't already know k? For instance, how do you know k = 5 doesn't work?
-Dan
Let me make my question a bit clearer. You previously generated a list of a_n and b_n. What you seemed to be doing was specifying a series a_n(k) and b_n(k). For the series you gave k = 3. But if k = 6 you would not generate that same series. Yes, I know there isn't one, but you haven't shown it won't work that way.Albert said:gven :$x,y\in N$
to prove :$\dfrac {x^2+y^2+1}{xy}=k(constant)\in N---(1)$
from (1) we have:
$x^2-kxy+y^2+1=0----(2)$
if $x<y$
for the same value of $k$ ,the soltions of (1)=the solutions of (2) will be :
$x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\, y$
$a_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b_1$
$a_2=b_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b_2$
$a_3=b_2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b_3$
-----
-----
$a_n=b_{n-1}\,\,\,\,\,\,\,\,\,b_n$
for $n=1,2,3------, a_n=b_{n-1}<b_n$
and :
$y=b_{n-1}=a_n=\dfrac{kx+\sqrt {k^2x^2-4(x^2+1)}}{2}-----(*)$
for the same value of $k$ (*) must be all satisfied,and then we can find $k$
topsquark said:Let me make my question a bit clearer. You previously generated a list of a_n and b_n. What you seemed to be doing was specifying a series a_n(k) and b_n(k). For the series you gave k = 3. But if k = 6 you would not generate that same series. Yes, I know there isn't one, but you haven't shown it won't work that way.
-Dan
It's a good approach. I worked it out and it says that k must be divisible by 3. I'm going nuts to find a way to narrow that down to k = 3, but with little no real success...I can disprove some k values, but it's nothing like a proof. I like bacterius' solution (I think it's fixed now?), but your approach is closer to mine. Which is why I'm picking your proof apart...it might lead to something I can use in my work.Albert said:$x^2 \,\,mod \,\,3 =0 \,\,\,\, or \,\, 1$
$y^2 \,\,mod \,\,3 =0 \,\, \,\, or \,\, 1$
in this case 0 will be deleted,so we let :
$x=3p\pm 1$
$y=3q\pm 1$ ($p,q\in N \,\ and \, \,p<q$)
for we set $x<y$
this may give us a hint to prove $k=3$ is the only answer
I am thinking----
topsquark said:It's a good approach. I worked it out and it says that k must be divisible by 3. I'm going nuts to find a way to narrow that down to k = 3, but with little no real success...I can disprove some k values, but it's nothing like a proof. I like bacterius' solution (I think it's fixed now?), but your approach is closer to mine. Which is why I'm picking your proof apart...it might lead to something I can use in my work.
-Dan
[sp]greg1313 said:Let $x$ and $y$ be positve integers such that $xy$ divides $x^2+y^2+1$. Show that $$\frac{x^2+y^2+1}{xy}=3$$
A divisibility problem in number theory refers to determining whether one number is divisible by another number without leaving a remainder. This is often used in mathematics to simplify fractions and solve equations.
To check if a number is divisible by another number, you can use the division algorithm. This involves dividing the larger number by the smaller number and checking if the remainder is equal to 0. If it is, then the number is divisible by the other number.
Divisibility plays a crucial role in number theory as it helps in identifying patterns and relationships between numbers. It is also used to find common factors and multiples, which are essential in solving problems related to fractions, equations, and prime numbers.
A number is divisible by 2 if its last digit is even (0, 2, 4, 6, or 8). Similarly, a number is divisible by 5 if its last digit is either 0 or 5. For larger numbers, you can use the division algorithm to check for divisibility by 2 or 5.
Yes, a number can be divisible by both 2 and 3. In fact, any number that is divisible by both 2 and 3 is also divisible by their product, 6. This is because 6 is the smallest number that is divisible by both 2 and 3.