# Integer Solutions

• July 30th 2009, 08:07 PM
Haven
Integer Solutions
Help!

Show that $2*y^2$ = $x^4$- 17 has no integer solutions.
(Hint: Show using quadratic reciprocity that y is a square mod 17 by considering each prime
divisor of y. Use this to show 2 is a fourth power mod 17, a contradication.)

Using the fundamental theorem of arithmetic, i assume $y = {p_1}^{a_1} *....* {p_t}^{a_t}$

By Euler's Criterion (y/17) $\equiv$ $y^8$ (mod 17)

I get if any of the exponents are even, then $(\frac{{p_i}^{a_i}}{17}) = 1$
If any of the exponents are odd, then $(\frac{{p_i}^{a_i}}{17}) = (\frac{p_i}{17})$

This is where i get stuck. (Sorry about the bad notation, i haven't quite figured out LaTeX)
• July 31st 2009, 09:18 AM
PaulRS
Let $p$ be a prime dividing $y$. Then $
x^4 - 17 \equiv 0\left( {\bmod .p} \right)
$

That is: $
x^4 \equiv 17\left( {\bmod .p} \right)
$
so 17 is a quadratic residue module p. So it follows that $
\left( {\tfrac{{17}}
{p}} \right)_L = 1
$
$(*)$

By the Quadratic Reciprocity Theorem: $
\left( {\tfrac{p}
{{17}}} \right)_L \left( {\tfrac{{17}}
{p}} \right)_L = \left( { - 1} \right){}^{\tfrac{{p - 1}}
{2} \cdot \tfrac{{17 - 1}}
{2}} = 1
$
hence it follows that $
\left( {\tfrac{p}
{{17}}} \right)_L = 1
$
, that is, all primes dividing $y$ are quadratic residues module 17, hence so is $y$.

But clearly, reading the equation module 17 we have: $
2y^2 \equiv x^4 \left( {\bmod .17} \right)
$
since $
y \equiv z^2 \left( {\bmod .17} \right)
$
for some integer $z$, coprime to $17$, it follows that: $
2z^4 \equiv x^4 \left( {\bmod .17} \right) \Rightarrow 2 \equiv \left( {z^{ - 1} x} \right)^4 \left( {\bmod .17} \right)
$
which is impossible since $2$ is not the 4th power of an integer module 17. Thus we conclude that the equation can have no solutions.

- an easy way to check this it to remember that ( $a$ being coprime to p) $x^n\equiv _{p} a$ has solutions if and only if $
a^{\tfrac{{p - 1}}
{{\left( {p - 1,n} \right)}}} \equiv 1\left( {\bmod .p} \right)
$
, in our case n=4 and p = 17 and we see that $
2^{\tfrac{{16}}
{{\left( {16,4} \right)}}} = 2^4 \equiv - 1\left( {\bmod .17} \right)
$
-

$(*)$ The prime p cannot be 17, because $y$ cannot be divisible by 17, otherwise the equation would imply that 17 also divides $x$. Let $y=17\cdot y'$ and $x=17 \cdot x'$ then $
2y^2 = x^4 - 17 \Rightarrow 2\left( {y'} \right)^2 \cdot \left( {17^2 } \right) - \left( {x'} \right)^4 \cdot \left( {17^4 } \right) = - 17
$
and note that the LHS is dividible by 17² while the RHS is not.