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Find all prime solutions of the equation $5x^2-7x+1=y^2.$

It is easy to see that $y^2+2x^2=1 \mod 7.$ Since $\mod 7$-residues are $1,2,4$ it follows that $y^2=4 \mod 7$, $x^2=2 \mod 7$ or $y=2,5 \mod 7$ and $x=3,4 \mod 7.$

In the same way from $y^2+2x=1 \mod 5$ we have that $y^2=1 \mod 5$ and $x=0 \mod 5$ or $y^2=-1 \mod 5$ and $x=4 \mod 5.$

How put together the two cases?

Computer find two prime solutions $(3,5)$ and $(11,23).$

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  • $\begingroup$ $y^2 \equiv -1 \pmod 5$ and $x \equiv 1\pmod 5$, or $x = 5$ or $y = 5$ $\endgroup$
    – Doug M
    Commented Jan 27, 2017 at 22:28

5 Answers 5

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Since $(0,1)$ is a solution to $5x^2-7x+1=y^2$, it can be used to parametrize all rational solutions to $5x^2-7x+1=y^2$. That will give us:

$$x:=\frac{-2ab - 7b^2}{a^2 - 5b^2}$$ and $$y:=\frac{a}{b}x+1$$

where $a,b\in \mathbb{Z}$, $b\neq 0$.

Since $x$ is prime, it follows that either $b=1$ or $x=b$.

  • case $b=1$

    This gives us $x:=(-2a - 7)/(a^2 - 5)$. Since $x\in \mathbb{Z}$, we get $a=\pm 2$, and then $x=11$ or $x=3$. Those will give $y=23$ or $y=5$, repectively.

  • case $x=b$.

We have that $\frac{-2ab - 7b^2}{a^2 - 5b^2}=b$ gives

$$(*) \hspace{2cm} a(a + 2) =5b^2 - 7b.$$

Since $y=a+1$ and $x=b$ are prime, and $x=2$ or $y=2$ do not give solutions to $5x^2-7x+1=y^2$, we conclude that $y=a+1$ and $x=b$ are ODD primes. In particular, $a(a + 2)\equiv 0 \mod 4$. Now reducing (*) $\mod 4$, contradicts the fact that $b$ is odd. Thus, the case $x=b$ does not occur.

Therefore $(x,y)= (11,23)$ and $(x,y)=(3,5)$ are the only solutions where both coordinates are prime numbers.

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Completing the square and dividing by $5$, we have

$$ (10 x - 7)^2 - 20 y^2 = 29$$

Thus $z = 10 x - 7$ and $w = 2 y$ are solutions of the Pell-type equation

$$ z^2 - 5 w^2 = 29$$

The positive integer solutions of this can be written as

$$\pmatrix{z\cr w\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{7\cr 2\cr} \ \text{or}\ \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{23\cr 10\cr}$$ for nonnegative integers $n$.

Now you want $w$ to be even and $z \equiv 3 \mod 10$. All the solutions will have $w$ even, and $z$ altermately $\equiv \pm 3 \mod 10$. Thus for $n$ even you get integers for $x,y$ with $$ \pmatrix{z_n\cr w_n\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{23\cr 10\cr}$$ and for $n$ odd, $$ \pmatrix{z_n\cr w_n\cr} = \pmatrix{9 & 20\cr 4 & 9\cr}^n \pmatrix{7\cr 2\cr}$$ You do get primes for $n=0$ ($z_0 = 23, w_0 = 10, x_0 = 3, y_0 = 5$) and $n=1$ ($z_1 = 103, w_1 = 46, x_1 = 11, y_1 = 23$). In general,

  • $x_n \equiv 0 \mod 3$ for $n \equiv 0$ or $3 \mod 4$.
  • $x_n \equiv 0 \mod 17$ for $n \equiv 2$ or $3 \mod 6$.
  • $x_n \equiv 0 \mod 5$ or $y_n \equiv 0 \mod 5$ for $n \equiv 0,3, 4, 5, 6, 9 \mod 10$.
  • $x_n \equiv 0 \mod 11$ for $n \equiv 1, 8 \mod 10$.
  • $x_n \equiv 0 \mod 13$ for $n \equiv 3, 5, 6, 7, 8, 10 \mod 14$.
  • $y_n \equiv 0 \mod 23$ for $n \equiv 1, 2, 5, 6 \mod 8$.

And every integer $n$ is in at least one of these classes. We conclude there are no other prime solutions.

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Try working mod $3$ and mod $8$. Assuming $x, y>3$, we have $x,y = \pm 1$ mod $3$. Since $x, y$ are odd we have $x^2, y^2=1$ mod $8$, so $$x^2, y^2 = 1 \text{ mod } 24.$$ Substituting in the equation gives $$x = 24k+11 $$ for some integer $k$. Rearranging the original equation we get $$x(5x-7)=(y-1)(y+1), \tag{1}$$ therefore $x |y-1$ or $x|y+1$, since $x$ is a prime number.

Solving for $x$ gives $$ x = \frac{17}{10} + \frac{1}{10}\sqrt{20y^2+29}>\frac{1}{3}(y+1).$$ Note that $x$ is odd and $y \pm 1$ is even, so $x \ne y\pm1$. This forces $x = \frac{1}{2} (y \pm1)$, or $$y = 2x \pm 1 = 48k + 22 \pm 1 \Rightarrow y = 48k+23.$$ $48k+21$ is rejected being divisible by 3. Plugging these in $(1)$ gives the solution $k=0$ or $$x = 11, \space y = 23.$$

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  • $\begingroup$ I would add a repetition about the inequalities, $(y+1)/ 3 < x < y-1.$ I think you are also rejecting $48 k + 21$ based on being divisible by $3$ $\endgroup$
    – Will Jagy
    Commented Jan 27, 2017 at 23:18
  • $\begingroup$ Yes, but I think they are obvious enough, aren't they? $\endgroup$
    – Emax
    Commented Jan 27, 2017 at 23:25
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Can we just charge straight at this?

$y$ is odd. $x=2 \ (\Rightarrow y^2=7)$ is not a solution, so $x$ is an odd prime.

$x(5x-7) = (y-1)(y+1)$, so $x \mid (y-1) $ or $x \mid (y+1)$ ($x$ is prime) so $kx=y\pm1$, $k$ even

$k\ge4$ is too large: $(kx\pm1)^2\ge (4x-1)^2 $ $= 16x^2-8x+1$ $>5x^2-7x+1$. So only $k=2$, that is $x=\frac 12(y\pm1)$, makes the equality feasible.

Considering the two cases:

  • (1) $x=\frac 12(y+1)$, $y=2x-1$:
    $x(5x-7) = 4x(x-1) \implies x = 3, y=5$

  • (2) $x=\frac 12(y-1)$, $y=2x+1$:
    $x(5x-7) = 4x(x+1) \implies x = 11, y=23$

Note that I didn't constrain $y$ at any point - the two solutions just happened to have $y$ prime.

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  • $\begingroup$ why the multiplication is not fleasible if, say, $x=\frac{1}{3}(y \pm 1)?$ $\endgroup$
    – Leox
    Commented Jan 28, 2017 at 8:09
  • $\begingroup$ Ок, then what about $x=\frac{1}{4}(y \pm 1)?$ $\endgroup$
    – Leox
    Commented Jan 28, 2017 at 16:32
  • $\begingroup$ well, you should prove that any $k>4$ is bad $\endgroup$
    – Leox
    Commented Jan 28, 2017 at 18:12
  • $\begingroup$ it's not obviously $\endgroup$
    – Leox
    Commented Jan 28, 2017 at 19:03
  • $\begingroup$ @Leox another update to answer $\endgroup$
    – Joffan
    Commented Jan 28, 2017 at 19:11
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If there is a solution it has more than 4000 digits, which makes me think there is no solution beyond the one two already mentioned.

In Mathematica,

i=1;
ans=Solve[5 x ^2-7x+1==y^2&&x>0&&y>0,{x,y},Integers];
ans2=ans/.C[1]->i//RootReduce
Dynamic@i
Dynamic[IntegerLength@ans2[[All,All,2]]]

and then run

    While[FreeQ[PrimeQ[ans2[[All,All,2]]],{True,True}],i++;ans2=ans/.C[1]->i//RootReduce]

Setting i=0 will stop the loop with the two known solutions:

{{x->3,y->5},{x->11,y->23}}

but there appear to be no easily found solutions beyond i=1

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