t's probably common knowledge that there are Diophantine equations which do not admit any solutions in the integers, but which admit solutions modulo nn for every nn. This fact is stated, for example,


**Conjecture** Let $m,n $be integer,find all
$$m^4+n^4=10m^2n^2+1$$


I this equality have only foursolution$(m,n)=(0,1),(0,-1),(1,0),(-1,0)$.But I can't prove it



Solve this diophantine equation: $m^4+n^4=10m^2n^2+1$
You have already listed all the possible solutions $(m,n)$ in which either $m=0$ or $n=0$.


Let us suppose that $(m,n) \in \mathbb{N} \times \mathbb{N}$ is a solution of your equation. Then, the discriminant $\Delta$ of the polynomial


$$p(z)=z^{2} - 10 \, z \,n^{2} +(n^{4}-1)$$


is necessarily a perfect square. Since


$$\Delta = 100n^{4}-4(n^{4}-1) = 96n^{4}+4=4(24n^{4}+1)$$


and $\gcd(4,24n^{4}+1)=1$, it follows that


$$24n^{4}+1 = \ell^{2}$$


for some $\ell \in \mathbb{Z}^{+}$. Since any equation of the type $y^{2}=Dx^{4}+1$, where $D>0$ and is not a perfect square, has at most two solutions in positive integers (cf. L. J. Mordell, Diophantine equations, p. 270: if I understand correctly the said result is original with the industrious [Wilhelm Ljunggren](https://mathoverflow.net/questions/206645/on-a-result-attributed-to-w-ljunggren-and-t-nagell) who would also remark that "if the equation has two solutions in positive integers, these will be determined by the fundamental unit in $\mathbb{Q}(\sqrt{D})$ and its square, apart from the case $D=1785$, where the solutions are given by the first and the fourth power of the fundamental unit"), it follows that your equation has at most four different solutions in positive integers.


