Find all integer solutions to the equation $y^2=2x^4+17$.
Problem
Source: 2014 Thailand October Camp Number Theory Exam p2
Tags: number theory, Diophantine equation
Mr.Math.Boy
05.03.2022 18:16
Can we consider the $y^2-2(x^2)^2=17$ Pell equation!?
NTstrucker
05.03.2022 18:38
Mr.Math.Boy wrote:
Half of the solution:
On the basis of the QM-GM we can see that $\sqrt{\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+1}{4}}\ge \sqrt[4]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}=\sqrt{abc}\Leftrightarrow d\ge 4$
Zyz1010 wrote:
Sorry ,but it should be $\frac{a^2+b^2+c^2+1}{4}\geq \sqrt[4]{a^2b^2c^2}$.
Mr.Math.Boy wrote:
Hint:
Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$.
We can demonstrate by induction and NZQR method, that $g(x)=x+1$.
PS: I sendt you a PM with the complete solution.
pco wrote:
You forgot $g(x)\equiv 1$
Mr.Math.Boy wrote:
Solution for non negative integers!
We can see that the $x=1, y=0$ it is solution. We will prove that we have not any other solution.
Indeed: we can see that we must have $y=3z$, so result ${{x}^{2}}-3{{z}^{2}}={{3}^{x+3z-2}}$.
We can see that we must have $x=3t$, so result $3{{t}^{2}}-{{z}^{2}}={{3}^{x+3z-2}}$.
In this way we can continue until that the exponent of the RHS will be negative, so the RHS it is $1/3^s$, but the LHS well be integer, contradiction.
Tip_pay wrote:
Find all integer solutions of the equation $(x,y)$
$$3x^2-y^2=3^{x+y}$$and prove that there are no others
Mr.Math.Boy wrote:
The Pell equation $a^2-2b^2=17$ have not integer solution, because taking the modulo $3$ we get $3A+p^2-2q^2=15+2$ with $p,q\in \left\{ 0,1 \right\}$ it is impossible
test20 wrote:
This does not seem to be true at all:
Take for instance $a=5$ and $b=2$.
If you are not sure about your ideas, please be silent and patient rather than hastily posting wrong solutions. To below: Mr.Math.Boy wrote: These solutions all there are correct! Think before you write untruth! Why are you lying through your teeth? (links and reference added.)
Mr.Math.Boy
05.03.2022 18:42
NTstrucker wrote: Mr.Math.Boy wrote: Hint: Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$. We can demonstrate by induction and NZQR method, that $g(x)=x+1$. PS: I sendt you a PM with the complete solution. Mr.Math.Boy wrote: Solution for non negative integers! We can see that the $x=1, y=0$ it is solution. We will prove that we have not any other solution. Indeed: we can see that we must have $y=3z$, so result ${{x}^{2}}-3{{z}^{2}}={{3}^{x+3z-2}}$. We can see that we must have $x=3t$, so result $3{{t}^{2}}-{{z}^{2}}={{3}^{x+3z-2}}$. In this way we can continue until that the exponent of the RHS will be negative, so the RHS it is $1/3^s$, but the LHS well be integer, contradiction. [/hide] If you are not sure about your ideas, please be silent and patient rather than hastily posting wrong solutions. These solutions all there are correct! Think before you write untruth! Are you sure that the this hint it is wrong? Mr.Math.Boy wrote: Can we consider the $y^2-2(x^2)^2=17$ Pell equation!? Can you prove that I am wrong? Have you a better hint for this problem?
rabbit317
05.03.2022 18:51
So, uh, try a little parity and find that $y^2$ must be odd and so $y$ must be odd.
Mr.Math.Boy
05.03.2022 19:01
The Pell equation $a^2-2b^2=17$ have not integer solution, because taking the modulo $3$ we get $3A+p^2-2q^2=15+2$ with $p,q\in \left\{ 0,1 \right\}$ it is impossible
kent2207
05.03.2022 19:03
Can I have the source about NZQR method? NTstrucker wrote:
Mr.Math.Boy wrote:
Half of the solution:
On the basis of the QM-GM we can see that $\sqrt{\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+1}{4}}\ge \sqrt[4]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}=\sqrt{abc}\Leftrightarrow d\ge 4$
Mr.Math.Boy wrote:
Hint:
Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$.
We can demonstrate by induction and NZQR method, that $g(x)=x+1$.
PS: I sendt you a PM with the complete solution.
Mr.Math.Boy wrote:
Solution for non negative integers!
We can see that the $x=1, y=0$ it is solution. We will prove that we have not any other solution.
Indeed: we can see that we must have $y=3z$, so result ${{x}^{2}}-3{{z}^{2}}={{3}^{x+3z-2}}$.
We can see that we must have $x=3t$, so result $3{{t}^{2}}-{{z}^{2}}={{3}^{x+3z-2}}$.
In this way we can continue until that the exponent of the RHS will be negative, so the RHS it is $1/3^s$, but the LHS well be integer, contradiction.
If you are not sure about your ideas, please be silent and patient rather than hastily posting wrong solutions.
Mr.Math.Boy
05.03.2022 19:18
kent2207 wrote: Can I have the source about NZQR method? NTstrucker wrote:
Mr.Math.Boy wrote:
Half of the solution:
On the basis of the QM-GM we can see that $\sqrt{\frac{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+1}{4}}\ge \sqrt[4]{{{a}^{2}}{{b}^{2}}{{c}^{2}}}=\sqrt{abc}\Leftrightarrow d\ge 4$
Mr.Math.Boy wrote:
Hint:
Let $g(x)=f(x)+1$, we get $g(xy)+g(x+y)=g(x)g(y)+1$.
We can demonstrate by induction and NZQR method, that $g(x)=x+1$.
PS: I sendt you a PM with the complete solution.
Mr.Math.Boy wrote:
Solution for non negative integers!
We can see that the $x=1, y=0$ it is solution. We will prove that we have not any other solution.
Indeed: we can see that we must have $y=3z$, so result ${{x}^{2}}-3{{z}^{2}}={{3}^{x+3z-2}}$.
We can see that we must have $x=3t$, so result $3{{t}^{2}}-{{z}^{2}}={{3}^{x+3z-2}}$.
In this way we can continue until that the exponent of the RHS will be negative, so the RHS it is $1/3^s$, but the LHS well be integer, contradiction.
If you are not sure about your ideas, please be silent and patient rather than hastily posting wrong solutions. The essence of the NZQR is the following: This method it is used for solving the functional equations. For example the $f(x+y)=f(x)+f(y)$ Cauchy equation: 1) We prove first, hat $f(x)=x$ or all $x$ from $N$. 2) After this, we prove the $f(x)=x$ for $x$ from $Z$. 3) After this, we prove the $f(x)=x$ for $x$ from $Q$. 4) An in final, using eventually the continuity, monotony, etc. we will prove (if we can) that $f(x)=x$ for all $x$ from $R$.
Math4Life2020
05.03.2022 19:40
(2,7) and (4,23). It doesn't seem like there are any others, though...
test20
05.03.2022 20:17
Mr.Math.Boy wrote: The Pell equation $a^2-2b^2=17$ have not integer solution, because taking the modulo $3$ we get $3A+p^2-2q^2=15+2$ with $p,q\in \left\{ 0,1 \right\}$ it is impossible This does not seem to be true at all: Take for instance $a=5$ and $b=2$.
grupyorum
05.03.2022 21:26
I can show $y\equiv 7\pmod{16}$ necessarily: Note that if $x$ is odd, then $y^2\equiv 3\pmod{8}$, absurd. Hence, $x$ is even and $y^2\equiv 17\equiv 49\pmod{32}$: $32\mid (y-7)(y+7)$. Note that as ${\rm gcd}(y-7,y+7)\mid 14$, it follows that $y\equiv \pm 7\pmod{16}$. Assume $y\equiv -7\pmod{16}$, set $y=16k-7$. Then $y^2-16 = (y-4)(y+4)=(16k-11)(16k-3)=2x^4+1$. Now, if $p\mid 2x^4+1$ a prime, then $(-2/p)=1$, thus $p\equiv 1,3\pmod{8}$. Notice, however, that $16k-3\equiv 5\pmod{8}$, thus it has a prime divisor of form $8i+5$ or $8j+7$, a contradiction. Hence, $y\equiv 7\pmod{16}$.
Mr.Math.Boy
05.03.2022 21:53
Mr.Math.Boy wrote: Can we consider the $y^2-2(x^2)^2=17$ Pell equation!? I solved this equation, and I get this: ${{x}^{2}}=\frac{1}{2\sqrt{2}}\left[ \left( 5+2\sqrt{2} \right){{\left( 3+2\sqrt{2} \right)}^{n}}-\left( 5-2\sqrt{2} \right){{\left( 3-2\sqrt{2} \right)}^{n}} \right]$
Slave
06.03.2022 07:38
Mr.Math.Boy wrote: Mr.Math.Boy wrote: Can we consider the $y^2-2(x^2)^2=17$ Pell equation!? I solved this equation, and I get this: ${{x}^{2}}=\frac{1}{2\sqrt{2}}\left[ \left( 5+2\sqrt{2} \right){{\left( 3+2\sqrt{2} \right)}^{n}}-\left( 5-2\sqrt{2} \right){{\left( 3-2\sqrt{2} \right)}^{n}} \right]$ But the $\text{RHS}$ is not necessary perfect square, so this gives nothing.
Mr.Math.Boy
06.03.2022 09:02
Yes, exactly this is the problem if the RHS can be or not perfect second power! Are you sure that this canot be happened? Can you prove this?
nnosipov
10.03.2022 17:23
Quidditch wrote: Find all integer solutions to the equation $y^2=2x^4+17$. I suppose that this is typos. The correct version may be $y^4=2x^2+17$ (and no integer solutions as an answer).
Quidditch
10.03.2022 19:30
nnosipov wrote: Quidditch wrote: Find all integer solutions to the equation $y^2=2x^4+17$. I suppose that this is typos. The correct version may be $y^4=2x^2+17$ (and no integer solutions as an answer). https://www.infinitydots.org/testarchive/tha_oct_2014_en.pdf That's the contest link so I don't know whether they made a typo or is it actually the problem...
kent2207
15.03.2022 13:56
Mr.Math.Boy wrote: [/hide] If you are not sure about your ideas, please be silent and patient rather than hastily posting wrong solutions.[/quote][/quote] The essence of the NZQR is the following: This method it is used for solving the functional equations. For example the $f(x+y)=f(x)+f(y)$ Cauchy equation: 1) We prove first, hat $f(x)=x$ or all $x$ from $N$. 2) After this, we prove the $f(x)=x$ for $x$ from $Z$. 3) After this, we prove the $f(x)=x$ for $x$ from $Q$. 4) An in final, using eventually the continuity, monotony, etc. we will prove (if we can) that $f(x)=x$ for all $x$ from $R$.[/quote] Ok thank you very much