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How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.

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How about using $\mod 5$? – Inceptio Mar 24 '13 at 7:33
Do you know the theorem of Fermat stating that no right-angled triangle has square area? – awllower Mar 24 '13 at 7:42

$$x^4-4y^4=z^2\iff\left(2y^2\right)^2+z^2=\left(x^2\right)^2.$$ Therefore if a primitive solution exists then $2y^2=2nm, \ z=m^2-n^2, \ x^2=m^2+n^2.$

From $2y^2=2nm\Rightarrow n=k^2,m=\ell^2$.
Therefore $x^2=\ell^4+k^4$ which does not have integer solutions (see this).

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Please explain the downvotes? Is my answer wrong? – P.. Mar 24 '13 at 7:47
I can be wrong, but aren't these just primitive Pythagoras triplets? You also can lose the needed invariant about positive arguments. – Harold Mar 24 '13 at 7:49
Thanks @awllower – P.. Mar 24 '13 at 21:13

Let show that the equation in question is equivalent to another one:$x^4+y^4=z^2$, called as $\Gamma$-equation.
If $p$ and $q$ and $r$ satisfy $(\Gamma)$, then, upon setting $x=r$, $y=pq$, and $z=p^4-q^4$, we obtain a solution of $x^4=4y^4+z^2$. Conversely, if $x$, $y$, and $z$ satisfy the equation in question, then there are $m$ and $n$ such that $x^2=m^2+n^2$, $2y^2=2mn$, and $z=m^2-n^2$. Hence we find that there are $p$ and $q$ with $m=p^2$ and $n=q^2$. Then $x^2=p^4+q^4$. Therefore the two equations are indeed equivalent, speaking in integers.
Now the transformed equation $(\Gamma)$ is easy to show its insolvability, and here we offer two methods:

I. Here have I proved that there is no pythagorean triple $(a,b,c)$ with both $a$ and $b$ squares, nor both $a$ and $c$ squares. If $(\Gamma)$ has a solution $(x,y,z)$ then $(x^2,y^2,z)$ is a pythagorean triple contradiction what is just said. So there is no solution at all.

II. Here is a proof contained in Diophantine Equation by Mordell.
Suppose $(x,y,z)$ is a solution with $(x,y)=1$. Since $x$ and $y$ cannot both be odd by considerations of $2$-powers, we assume that $y$ is even, and $x$ is odd. Now by Euclid's lemma, we know that there are $a$ and $b$ such that $x^2=a^2-b^2$, $y^2=2ab$, and $z=a^2+b^2$, with $a, b$ coprime. By dint of the same lemma again, we find $p$ and $q$ such that $x=p^2-q^2$, $b=2pq$, $a=p^2+q^2$, with $p, q$ coprime. Since $p$, $q$ and $p^2+q^2$ are pairwise prime to each other, and for $y^2=2ab=4pq(p^2+q^2)$, there are coprime $r, s$ and $t$ such that $p=r^2$, $q=s^2$, and $p^2+q^2=t^2$. So $r^4+s^4=t^2$, while $z=a^2+b^2=p^4+q^4+6p^2q^2=r^8+s^8+6r^4s^4$. Hence $z>(r^4+s^4)^2=t^4$, namely, $t$ is strictly less than $z$. So this completes the proof "by descent".
Though the length is still a pain, I hope that, this time, length brings not only pain, but the result, the richness of the subject, and its depth, as well. And tell me if something occurs that is inappropriate here. Thanks in advance.

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This proof reflects the underlying principle of many of elementary Diophantine equations that if we have an equation of the form $PQ=z^2$, where $P$ and $Q$ are polynomials, then $P$ and $Q$ are almost squares, i.e. they differ by some g.c.d. at most. Almost all statements found in Fermat's times could be solved by this principle. And André. Weil actually wrote a thesis on this principle, translated to the language of birational geometry, as his dotoral thesis. – awllower Mar 24 '13 at 8:17
Hope this could clarify some things and some doubts. – awllower Mar 25 '13 at 16:21

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