# Thread: Another Diophantine Question

1. ## Another Diophantine Question

Hello community. Recently, I received a homework problem that has completely stumped me.

Show that the Diophantine equation: x^4 - 2y^4 = 1 has no nontrivial solutions. My hint is to consider congruences modulo 5.

Thank you for any assistance you may offer.

2. Originally Posted by qwath
Hello community. Recently, I received a homework problem that has completely stumped me.

Show that the Diophantine equation: x^4 - 2y^4 = 1 has no nontrivial solutions. My hint is to consider congruences modulo 5.

Thank you for any assistance you may offer.
To consider congruences modulo 5 is to consider all the remainders,
$\displaystyle 5k$
$\displaystyle 5k+1$
$\displaystyle 5k+2$
$\displaystyle 5k+3$
$\displaystyle 5k+4$
When we raise to the exponent of 4 the forms take respectively,
$\displaystyle 5k$
$\displaystyle 5k+1$
$\displaystyle 5k+1$
$\displaystyle 5k+1$
You might have realized that is Fermat's Little Theorem.

But anyways the two possible forms of $\displaystyle a^4$ (a fourth power) are: $\displaystyle 5k,5k+1$.

Thus, the possibilities are in the Diophantine Equation:
$\displaystyle (5k)+2(5j)=5\cdot 0+1$---> Impossible, RHS different form.
$\displaystyle (5k)+2(5j+1)=5\cdot 0+1$---> Impossible
$\displaystyle (5k+1)+2(5j)=5\cdot 0+1$---> Possible
$\displaystyle (5k+1)+2(5j+1)=5\cdot 0+1$---> Impossible.

Thus, what your homework was asking was not correct.
The second case, is still possible.