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This question is an exact duplicate of the question

Does the equation $x^4+y^4+1=z^2$ have a non-trivial solution?

posted by Tito Piezas III on math.stackexchange.com.

The background of this question is this: Fermat proved that the equation, $$x^4 +y^4=z^2$$
has no solution in the positive integers. If we consider the near-miss, $$x^4 +y^4-1=z^2$$
then this has plenty (in fact, an infinity, as it can be solved by a Pell equation). But J. Cullen, by exhaustive search, found that the other near-miss, $$x^4 +y^4 +1=z^2$$
has none with $0 < x,y < 10^6$ .

Does the third equation really have none at all, or are the solutions just enormous?

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The question was also posted in January at math.stackexchange.com/questions/16887/… –  Tapio Rajala Apr 15 '11 at 9:25
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There are certainly plenty of rational solutions. Elkies (Math. Comp. vol. 51, no. 184 (1988), pp. 825&ndash;835) showed that the equation $A^4+B^4+C^4=D^4$ has infinitely many integer solutions; by a well-known theorem we can't have $C=0$, so then $(A/C,B/C,D^2/C^2) gives a rational solution to the equation above. –  Martin Bright Apr 15 '11 at 11:11
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Infiniteness of rational solutions is easy to obtain from the identity $$(xy)^4 + (xz)^4 + (yz)^4 = (z^4 - x^2y^2)^2$$ which holds for Pythagorean triples $(x,y,z)$ with $x^2+y^2=z^2$. –  Max Alekseyev Apr 20 '11 at 12:26
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Why is my question (asked in a Mathstackexchange post in Jan 2011) copied word-for-word by this poster? –  Tito Piezas III Aug 28 '13 at 3:50
20  
Link to 2013 meta.MO post about this apparent case of plagiarism: meta.mathoverflow.net/questions/729/… (Please vote up for visibility) –  Yemon Choi Aug 28 '13 at 4:09
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2 Answers

Didn't find any solutions with $1 \leq y \leq 7.9 \cdot 10^7$ and no restriction on $x,z$ in about 17 hours on 1 core.

Here is the search:

Per several discussions and arguments $\mod {20}$ both $x,y$ are divisible by $10$.

Rewrite as $$ y^4 + 1 = z^2 - x^4 $$

The RHS is a difference of two squares with the additional restriction the second square must be fourth power, so the algorithm uses a single loop, write $(10y_1)^4+1$ as a difference of two squares (in all possible ways) and checks for fourth power.

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Let take a look to the special case: $$ y=ax $$ where $a$ is a fixed nonzero integer. When $a=1$ was already observed by Max Alexseyev that there is no solution. Since
$$ x^4+y^4+(x+y)^4=2(x^2+xy+y^2)^2 $$ the equation $x^4+y^4+1=z^2$ can then be written: $$ X^2 -DY^4 =1 $$ with $$ X=z, Y=x, D = a^4+1. $$

(Or, more simply, we get this also directly from the original equation...).

Observe now that the equation $$ a^4+1=w^2 $$ has no integer solution $w$. See e.g., Corollary in page 17 of Mordell's Diophantine Equations.

Then it follows from the paper of Togbe et al. below that there are at most $2$ positive solutions $x,z$ of the equation.

Togbe, A.; Voutier, P. M.; Walsh, P. G.(3-OTTW) Solving a family of Thue equations with an application to the equation $x^2-Dy^4=1$. Acta Arith. 120 (2005), no. 1, 39–58. 11D59 (11D25)

Let $D$ be a positive nonsquare integer. The authors study the Diophantine equation $X^2-DY^4=1$ in positive integers $X$ and $Y$ and refine a theorem of W. Ljunggren [Skr. Norske Vid.-Akad. Oslo I 1936, no. 12, 1--73; Zbl 0016.00802]. Let $(T_1,U_1)$ be the smallest integer solution to the Pell equation $X^2-DY^2=1$. For $k\ge 1$, let $T_k+U_k\sqrt D={(T_1+U_1\sqrt D)^k}$ represent all positive integer solutions to the Pell equation. The authors prove: There are at most two positive integer solutions $(X,Y)$ to the equation $X^2-DY^4=1$. If two solutions $Y_1 <Y_2$ exist, then $Y_1^2=U_1$ and $Y_2^2=U_2$, except only if $D=1785$ or $D=16\cdot 1785$, in which case $Y_1^2=U_1$ and $Y_2^2=U_4$. If only one positive integer solution $(X,Y)$ exists, then $Y^2=U_l$ where $U_1=lv^2$ for some squarefree integer $l$, and either $l=1$, $l=2$, or $l=p$ for some prime $p\equiv 3\pmod 4$. The problem is reduced to solving the family of Thue equations $x^4+4tx^3y-6tx^2y^2-4t^2xy^3+t^2y^4=t_0^2$, where $t_0$ divides $t$ and $t_0\le \sqrt t$, for a positive integer $t$. However, it is not required to solve this family completely, but only for solutions whose quotient $x/y$ is near to $\beta^{(3)}$ or $\beta^{(4)}$, where $\beta^{(j)}$, $j=1,\dots,4$, denote the roots of the univariate polynomial corresponding to the Thue equation in a particular order defined in the paper. For these two roots, an effective measure of irrationality can be proved by Thue's hypergeometric method. Reviewed by Clemens Heuberger

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It looks like we always have $T_1=2a^4+1$ and $U_1=2a^2$. Thus $U_1$ is not a square, and we further have $l=2$ and $U_l=4a^2(2a^4+1)$ which is neither a square. Hence, there are no integer solutions to $X^2−(a^4+1)Y^4=1$. –  Max Alekseyev Apr 20 '11 at 21:10
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