Modulo $8$ seems to work fine.

If $y>=2$ then with mod8 we have $z=even$ $4^y=(5^k-x)(5^k+x)$ Now $(5^k-x,5^k+x)=2$ so we have to cases: Case1: $5^k-x=2$ $5^k+x=2^(2y-1)$ Add this two we have :. $5^k=1+2^(2y-2)$ By zigmondy theory no solution for $x>=5$ (other wise take mod8 and we have $k=evene$ and then square different) So a solution is :$(x,y,z):(3,2,2)$ Now if $y=1$ then we have: $x^2+4=5^z$ If $z=even$ obvious no solution We work in $Z[i]$. $(x-2i)(x+2i)=5^z$. Let $(x-2i,x+2i)=d$ then: $d|4$,$d|2x$ so $N(d)|16$ and $N(d)|4x^2$ and because $x=odd$ we have $N(d)|4$. On the other hand $N(d)|(x^2+4)^2$ so $N(d)|x^4$ and because $x=odd$ we have $N(d)=1$ so $d$ is a unit.

rafaello wrote: Modulo $8$ seems to work fine. Well, the hard case is when $y=1$. Indeed, then the problem becomes $5^z-x^2=4$ which is an old problem that can be solved easily via factorization in $\mathbb{Z}[i]$. See here for a previous discussion on this forum.

Tintarn wrote: rafaello wrote: Modulo $8$ seems to work fine. Well, the hard case is when $y=1$. Indeed, then the problem becomes $5^z-x^2=4$ which is an old problem that can be solved easily via factorization in $\mathbb{Z}[i]$. See here for a previous discussion on this forum. I think it is not very easy to prove $|a_m| = 2$ only holds for $m = 1, 3$. In fact, none of the proposers succeeded to solve the problem using factorization under $\mathbb{Z}[i]$.

$p^k=a^2+b^2$ can be represented uniquely. Upd: İf (a,b)=1 of course

rightways wrote: $p^k=a^2+b^2$ can be represented uniquely. Nope. $5^3=5^2+10^2=2^2+11^2$.

Tintarn wrote: rightways wrote: $p^k=a^2+b^2$ can be represented uniquely. Nope. $5^3=5^2+10^2=2^2+11^2$. But if we furthermore require that $p\nmid a,b$, then the representation is indeed (more-or-less) unique. ltf0501 wrote: Tintarn wrote: rafaello wrote: Modulo $8$ seems to work fine. Well, the hard case is when $y=1$. Indeed, then the problem becomes $5^z-x^2=4$ which is an old problem that can be solved easily via factorization in $\mathbb{Z}[i]$. See here for a previous discussion on this forum. I think it is not very easy to prove $|a_m| = 2$ only holds for $m = 1, 3$. In fact, none of the proposers succeeded to solve the problem using factorization under $\mathbb{Z}[i]$. Seconding this. Since the discriminant of the characteristic polynomial is negative, I am not aware of any easy way to bound the growth rate of the sequence. I would be happy to see one if you have one in mind. The only quantitative method I know of involves something non-elementary like Roth's, but at the same time I am not an expert at this and have only done a minimum amount of research......

You are right. I am not sure what exactly I had in mind back when I wrote the post in the other thread, but certainly it is not obvious to be me now why it should be true (probably I just miscalculated stuff).

rightways wrote: $p^k=a^2+b^2$ can be represented uniquely. Upd: İf (a,b)=1 of course OK, you are right, but how does it help to solve the problem?

@below corrected now

MrOreoJuice wrote:

Taking mod 3 we don't have $z=even$

I got carried by a few friends here. Solved with islander7, nukelauncher, tigershark22, Eyed, and me. (Also solved with someone who I'm not sure how to credit currently). First, we resolve $y$ being greater than 1. In this case, we get that modulo $8$, \[5^z\equiv x^2\equiv 1\pmod 8\]which implies $z$ is even. Letting $z=2z'$, we get \[(5^{z'}-x)(5^z+x)=2^{2y}\]Thus, we can compute that \[\gcd(5^{z'}-x,5^z+x)\mid\gcd(2\cdot 5^{z'}, 2^{2y})=2\]so thus $5^{z'}-x=2$ and $5^{z'}+x=2^{2y-1}$. This means that \[5^{z'}=1+2^{2y-2}\]which will die to either Mihăilescu's theorem or Zsigmondy, whichever the reader prefers, leaving the only solution as $(x,y,z')=(3,2,1)$ or $(x,y,z)=(3,2,2)$. Onto the harder case. We can first note $z$ must be odd, as \[5^z\equiv 4+x^2\equiv 5\pmod 8\]Now, in this case, we apply the standard negative Pell equation \[5k^2-4=x^2\]However, this follows from a much more general result: Claim. A number $k$ is a Fibonacci number if and only if $5k^2\pm 4$ is a perfect square. Proof. To prove this, consider the Lucas and Fibonacci sequences, $\{L_j\}_{j\geq 0}$ and $\{F_j\}_{j\geq 0}$, respectively, such that \[L_0=2\qquad L_1=1\qquad L_{j+1}=L_j+L_{j-1}\]\[F_0=0\qquad F_1=1\qquad F_{j+1}=F_j+F_{j-1}\]Now, we quickly prove the following lemma: Lemma. \[(-1)^j+F_j^2=F_{j+1}F_{j-1}\]\[L_j=F_{j+1}+F_{j-1}\]Proof. For the first part, note that \[\begin{pmatrix}F_{j+1}\\F_j\end{pmatrix}=\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}F_j\\F_{j-1}\end{pmatrix}\]implying that \[\begin{pmatrix}F_{j+1}&F_j\\F_j&F_{j-1}\end{pmatrix}=\begin{pmatrix}1&1\\1&0\end{pmatrix}^n\]Taking the determinant finishes the first part off. To show the second part, we shall induct on $j$: Base Case. $j=1,2$. This can be done by manual computation. Induction Hypothesis. Assume the result is true for some $j=m,m-1\geq 1$. We show it for $j=m+1$. Induction Step. We can note \begin{align*} L_{j+1}&=L_j+L_{j-1}\\ &=F_{j+1}+F_j+F_{j-1}+F_{j-2}\\ &=F_{j+2}+F_j \end{align*}as desired. Now, to finish off the proof of one direction, we can note that \[L_j^2=F_{j+1}^2+F_{j-1}^2+2F_{j+1}F_{j-1}=4(-1)^j+(F_{j+1}-F_{j-1})^2+4F_j^2\]or \[5F_j^2+4(-1)^j=L_j^2\] For the other direction, we can note the convergents of $\phi\frac{1+\sqrt 5}2$ are indeed $\frac{F_{j+1}}{F_j}$. Now, assuming that $5k^2\pm 4=m^2$, we get that $\frac{m+n}2=\ell$ is an integer. Thus, we can write this as \[\ell^2-\ell n-n^2=\pm 1\]implying $\gcd(\ell,n)=1$ and that \[\pm\frac 1{n^2}=\left(\frac kn-\phi\right)\left(\frac kn+\phi^{-1}\right)\]implying that as \[\frac kn+\phi^{-1}>\frac 32+\frac 12=2\]we have \[\left|\frac kn-\phi\right|<\frac 1{2n^2}\]Thus $n$ must be Fibonacci, as $\frac kn$ is a convergent. $\blacksquare$ Now, we are almost done! We need to characterize Fibonacci numbers that are powers of $5$. This follows from the following magical fact: Claim. $\nu_5(F_k)=\nu_5(k)$. Proof. We note that $\sqrt 5$ is irrational, and thus will the following fact. Lemma. \[\frac{L_n+F_n\sqrt 5}2=\phi^n\]Proof. We can just induct on $n$: Base Case. $n=0,1$. This is true by simple arithmetic. Induction Hypothesis. Assume the result is true for $n=j,j-1$. We show it is true for $n=j+1$. Induction Step. As $\phi^2=\phi+1$, we get \begin{align*} \phi^{j+1}&=\phi^j+\phi^{j-1}\\ &=\frac{L_j+L_{j-1}+(F_j+F_{j-1})\sqrt 5}2\\ &=\frac{L_{j+1}+F_{j+1}\sqrt 5}2 \end{align*} Now, we can note that $\sqrt 5$ is irrational (no I will not prove this. I'm braindead after writing all of the above), so thus we can use the binomial theorem to get \[2^{n-1}F_n=\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom n{2k+1}5^k=n+\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\frac n{2k+1}\binom{n-1}{2k}5^k\]Now, to finish this entire problem off, we can note that $\nu_5(2k+1)<k$, so thus every number in the second sum is divisible by $5^{\nu_5(n)+1}$. Thus \[\nu_5(F_n)=\nu_5(2^{n-1}F_n)=\nu_5(n)\]completing the proof. $\blacksquare$ The rest of the problem is pretty easy, right? We must have equality, so $F_n=n$, which is absurd for $n\geq7$. Thus, we get $n=1,5$ (wait what was $n$ again) implying that the only solutions are $z=1$ and $z=3$, which lead to $(1,1,1)$ and $(11,1,3)$. I'm done (maybe?)! Here are some additional comments: I was actually working with $\mathbb Q(i)$, which seems to work fairly nicely for the general idea $x^2+4^y=w^z$. I don't think I'm near complete, but I do believe that it can solve the problem at hand. On immediate look, this problem (reduced to $y=1$) looks like a Lebesgue Diophantine, but turns out to be different (I won't say harder) as the standard techniques don't seem to immediately work on this.

naman12 wrote: I got carried by a few friends here. Solved with islander7, nukelauncher, tigershark22, Eyed, and me. (Also solved with someone who I'm not sure how to credit currently). First, we resolve $y$ being greater than 1. In this case, we get that modulo $8$, \[5^z\equiv x^2\equiv 1\pmod 8\]which implies $z$ is even. Letting $z=2z'$, we get \[(5^{z'}-x)(5^z+x)=2^{2y}\]Thus, we can compute that \[\gcd(5^{z'}-x,5^z+x)\mid\gcd(2\cdot 5^{z'}, 2^{2y})=2\]so thus $5^{z'}-x=2$ and $5^{z'}+x=2^{2y-1}$. This means that \[5^{z'}=1+2^{2y-2}\]which will die to either Mihăilescu's theorem or Zsigmondy, whichever the reader prefers, leaving the only solution as $(x,y,z')=(3,2,1)$ or $(x,y,z)=(3,2,2)$. Onto the harder case. We can first note $z$ must be odd, as \[5^z\equiv 4+x^2\equiv 5\pmod 8\]Now, in this case, we apply the standard negative Pell equation \[5k^2-4=x^2\]However, this follows from a much more general result: Claim. A number $k$ is a Fibonacci number if and only if $5k^2\pm 4$ is a perfect square. Proof. To prove this, consider the Lucas and Fibonacci sequences, $\{L_j\}_{j\geq 0}$ and $\{F_j\}_{j\geq 0}$, respectively, such that \[L_0=2\qquad L_1=1\qquad L_{j+1}=L_j+L_{j-1}\]\[F_0=0\qquad F_1=1\qquad F_{j+1}=F_j+F_{j-1}\]Now, we quickly prove the following lemma: Lemma. \[(-1)^j+F_j^2=F_{j+1}F_{j-1}\]\[L_j=F_{j+1}+F_{j-1}\]Proof. For the first part, note that \[\begin{pmatrix}F_{j+1}\\F_j\end{pmatrix}=\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}F_j\\F_{j-1}\end{pmatrix}\]implying that \[\begin{pmatrix}F_{j+1}&F_j\\F_j&F_{j-1}\end{pmatrix}=\begin{pmatrix}1&1\\1&0\end{pmatrix}^n\]Taking the determinant finishes the first part off. To show the second part, we shall induct on $j$: Base Case. $j=1,2$. This can be done by manual computation. Induction Hypothesis. Assume the result is true for some $j=m,m-1\geq 1$. We show it for $j=m+1$. Induction Step. We can note \begin{align*} L_{j+1}&=L_j+L_{j-1}\\ &=F_{j+1}+F_j+F_{j-1}+F_{j-2}\\ &=F_{j+2}+F_j \end{align*}as desired. Now, to finish off the proof of one direction, we can note that \[L_j^2=F_{j+1}^2+F_{j-1}^2+2F_{j+1}F_{j-1}=4(-1)^j+(F_{j+1}-F_{j-1})^2+4F_j^2\]or \[5F_j^2+4(-1)^j=L_j^2\] For the other direction, we can note the convergents of $\phi\frac{1+\sqrt 5}2$ are indeed $\frac{F_{j+1}}{F_j}$. Now, assuming that $5k^2\pm 4=m^2$, we get that $\frac{m+n}2=\ell$ is an integer. Thus, we can write this as \[\ell^2-\ell n-n^2=\pm 1\]implying $\gcd(\ell,n)=1$ and that \[\pm\frac 1{n^2}=\left(\frac kn-\phi\right)\left(\frac kn+\phi^{-1}\right)\]implying that as \[\frac kn+\phi^{-1}>\frac 32+\frac 12=2\]we have \[\left|\frac kn-\phi\right|<\frac 1{2n^2}\]Thus $n$ must be Fibonacci, as $\frac kn$ is a convergent. $\blacksquare$ Now, we are almost done! We need to characterize Fibonacci numbers that are powers of $5$. This follows from the following magical fact: Claim. $\nu_5(F_k)=\nu_5(k)$. Proof. We note that $\sqrt 5$ is irrational, and thus will the following fact. Lemma. \[\frac{L_n+F_n\sqrt 5}2=\phi^n\]Proof. We can just induct on $n$: Base Case. $n=0,1$. This is true by simple arithmetic. Induction Hypothesis. Assume the result is true for $n=j,j-1$. We show it is true for $n=j+1$. Induction Step. As $\phi^2=\phi+1$, we get \begin{align*} \phi^{j+1}&=\phi^j+\phi^{j-1}\\ &=\frac{L_j+L_{j-1}+(F_j+F_{j-1})\sqrt 5}2\\ &=\frac{L_{j+1}+F_{j+1}\sqrt 5}2 \end{align*} Now, we can note that $\sqrt 5$ is irrational (no I will not prove this. I'm braindead after writing all of the above), so thus we can use the binomial theorem to get \[2^{n-1}F_n=\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom n{2k+1}5^k=n+\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\frac n{2k+1}\binom{n-1}{2k}5^k\]Now, to finish this entire problem off, we can note that $\nu_5(2k+1)<k$, so thus every number in the second sum is divisible by $5^{\nu_5(n)+1}$. Thus \[\nu_5(F_n)=\nu_5(2^{n-1}F_n)=\nu_5(n)\]completing the proof. $\blacksquare$ The rest of the problem is pretty easy, right? We must have equality, so $F_n=n$, which is absurd for $n\geq7$. Thus, we get $n=1,5$ (wait what was $n$ again) implying that the only solutions are $z=1$ and $z=3$, which lead to $(1,1,1)$ and $(11,1,3)$. I'm done (maybe?)! Here are some additional comments: I was actually working with $\mathbb Q(i)$, which seems to work fairly nicely for the general idea $x^2+4^y=w^z$. I don't think I'm near complete, but I do believe that it can solve the problem at hand. On immediate look, this problem (reduced to $y=1$) looks like a Lebesgue Diophantine, but turns out to be different (I won't say harder) as the standard techniques don't seem to immediately work on this. Nice one! This is more or less the official solution. The only difference is that one direction of the key lemma can be carried out using some sort of infinite descent. I personally thought that this is an easy one when I reduce it to $y=1$ in like one minute. Ten minuted afterwards, I started to believe that there is not any elementary solution to this. Then I got mind-blown by the official solution.

One more additional comment (credits to one of the contestants and also my laptop lol): Using the fact that $\mathbb{Z}[i]$ is a UFD would give us quite quickly for the $y=1$ case that \[x+2i = \pm(1\pm 2i)^{2n+1}\]where $z=2n+1$. The first $\pm$ is redundant if we allow for negative $x$. Now we know that for the $+$ case, $n=0$ is a solution, and for the $-$ case, $n=1$ is a solution. It is in general hard to show that there are no other solutions as if $x_n+y_ni=(1\pm 2i)^{2n+1}$, then $y_n$ satisfies a second-order linear recurrence with negative discriminant, making bounding the terms extremely hard. However, one of the contestants made the following observation: Lemma. If $x_n+y_ni=(-3+4i)^n$ where $x_n,y_n\in \real$, then $v_2(x_n-1)=v_2(y_n)$. This can be proved once one also observed that $v_s(x_n-1)=v_2(y_n)=v_2(n)+2$ (credits to my laptop ) and induct on $v_2(n)$. With this lemma, now everything is pretty straightforward. If $x+2i=(1+2i)^{2n+1}$, then \[(1+2i)\left(-3+4i)^n-1\right) = x-1, \Rightarrow (-3+4i)^n-1 = a(1-2i)\]for some integer $a$. This is a contradiction to the lemma unless $a=0$ and $n=0$. If $x+2i=(1-2i)^{2n+1}$, then \[(-11+2i)\left((-3+4i)^{n-1}-1\right)=x-11,\Rightarrow (-3+4i)^{n-1}-1=a(-11-2i)\]for some integer $a$. This is once again a contradiction unless $a=0$ and $n=1$. There are plenty of properties that one can produce with this cool lemma. I think one could probably produce a bunch of extremely hard Diophantine equation using this mechanism (evil laugh).

This is not required now but anyways. Let $k\equiv 1 \pmod{2}$.We solve the equation $$x^{2}+4^{y}=(k^{2}-4)^{z}$$First let's handle the case if $y\geq 2$. Clearly $z$ is even in this case so one can just use zsigmondy theorem as above users have did to conclude that in this case solution only exists if $(k,x,y,z)=(3,3,2,2)$ For the case $y=1$, we have equation $$x^{2}+4=(k^{2}-4)l^{2}$$Now here comes our main claim. Claim If $\frac{x^2+4}{l^2}+4$ is a odd perfect square then $k=3$ Proof As done here just see $$\frac{x^2+4}{l^2}+4=k^2$$Which reduces to $$x^{2}+4=l^{2}(k^{2}-4)$$Now let $m=\frac{x+kl}{2}$ and the equation reduces to $m^{2}+l^{2}+1=l\cdot m\cdot k$ which by vieta jumping is known to have solutions only if $k=3$. $\blacksquare$ Also it's well known that the solutions to equation $m^{2}+l^{2}+1=l\cdot m\cdot 3$ are fibonacci numbers. Then we can continue the way naman12 did to end this.

My solution - (Will be happy if someone LaTeX this) If y >=2, then mod 2 arguments tells us that z is even, so if z = 2k, we have 4^y = 5^z - x^2 = (5^k)^2 - x^2 = (5^k-x)(5^k+x) If 5^k-x = 2, then the equation has only solution (x, y, z) = (3, 2, 2) due to Zsigmondy Theorem Let y = 1. Here we see that z must be odd due to mod 2 argument. sorry redact @below I am sorry friend I misunderstood Fermat Theorem statement

peterdawson wrote: So x^2 + 4 = 5^z imply x^2 + 5 = 5^z + 1, but if z is odd, then 3 | 5^z + 1 but due to Fermat Theorem on two square, a prime of form 4a+3 divide x^2 + k if and only if 4a+3 | x and 4a+3 |k but in this case it mean 3 divide 5 a contradiction. And what about the solution $x=11, z=3$? Note that $5$ on the LHS is not a perfect square, so how do you apply Fermat?

naman12 wrote: Now, we are almost done! We need to characterize Fibonacci numbers that are powers of $5$. Nice solution! Shorter finish from here: Just use the well-known and elementary fact that the gcd of two Fibonacci numbers is again Fibonacci (more precisely $(F_a,F_b)=F_{(a,b)}$). Now suppose $F_n=5^k$ with $k \ge 2$. Since $F_{25}=75025 =5^2 \cdot 3001$, their gcd is $25$ which is not Fibonacci. Contradiction! (This is a very general and useful trick that also works for numbers different from $5$.)

Not intended to be sarcastic, but is the title kind of troll?

NTstrucker wrote: Not intended to be sarcastic, but is the title kind of troll? LOL yeah the title is meant to be troll I think. The whole point is that this seemingly easy Diophantine equation is actually complicated, but at the same time not too complicated for an elementary solution to work. Lots of contestants were tricked by the appearance of the equation and got stuck on it though, so I guess it is a troll problem in this sense :p

Troll problem. Recognized key claim, but it took me quite some time to realize I'm done. The answer is $(1,1,1),(3,2,2), (11,1,3)$ and they obviously work. Now I show they are the only solutions. Case 1: $y>1$. In this case $z$ is even taking the equation mod $8$, call $2z'$. Then we get $(5^{z'}-2^y)(5^{z'}+2^y)$ is a perfect square. Since the two factors are coprime, they are squares themselves. Let $a^2=5^{z'}-2^y, b^2=5^{z'}+2^y$ then $(b+a)(b-a)=2^{y+1}$, so $b-a=2, b+a=2^y, b=2^{y-1}+1$. Hence $5^{z'}=b^2-2^y=(2^{y-1}+1)^2-2^y=4^{y-1}+1$. We can use LTE and size to deduce the only solution is $(x,y,2z')=(3,2,2)$. Case 2: $y=1$ Then we obtain $x^2+4=5^z$. Consider the two positive solutions of $a^2+b^2=5^z$. To prove there are two, note $N(a+bi)=N((1+2i)^z)$. Since $1+2i, 1-2i, 2+i, 2-i$ are primes in $\mathbb{Z}[i]$, we can see any solution to $a^2+b^2=5^z$ has $a+bi = (1+2i)^{c_1} (1-2i)^{c_2} (2+i)^{c_3} (2-i)^{c_4}$. Note if we use $1+2i$ we can't use $1-2i, 2+i$ since they would make the real part divisible by 5. Furthermore $i(2-i)=1+2i$ so we assume $a+bi=(1+2i)^{c_1}$ as swapping and negating $a,b$ don't matter. Since the case $z$ even has been solved, we consider $z$ odd. Starting with $1+2i$, we multiply by $(1+2i)^2=-3+4i$ each move. We casework on the sign of the imaginary part. Case 1: $(1+2i)(-3+4i)^k=(1+2i)(x+yi)=M+2i$. Then $(1+2i)((x-1)+yi))=M-1$, so $(-3+4i)^k-1=(x-1)+yi = \frac{M-1}{5} (1-2i)$. Suppose $(-3+4i)^k=s_k+t_ki$. I claim $\nu_2(s_k-1)=\nu_2(t_k)=2+\nu_2(k)$, which would contradict the above.

We induct on $l$, with the base case, $l=0$ being immediate. Note $(-3+4i)^{2^l}=((-3+4i)^{2^{l-1}})^2=(s_{2^{l-1}}+t_{2^{l-1}}i)^2$ We can see $s_{2^l}-1=s_{2^{l-1}}^2 - 1 - t_{2^{l-1}}^2$. Note $\nu_2(t_{2^{l-1}}^2)=2l+2, \nu_2(s_{2^{l-1}}^2 - 1)=\nu_2(s_{2^{l-1}}-1) + \nu_2(s_{2^{l-1}}+1)=1+l+1=l+2$. Since $l>0$, $\nu_2(s_{2^l}-1)=l+2$. $\nu_2(t_{2^l})=\nu_2(2s_{2^{l-1}}t_{2^{l-1}})=1+0+(l+1)=l+2$. Now, I prove $\nu_2(s_{2^l (2k+1)}-1)=\nu_2(s_{2^l}-1)$ and $\nu_2(t_{2^l (2k+1)})=\nu_2(t_{2^l})$ I will prove this with induction on $k$ alongside the previous claim. In the base case, $k=0$, there is nothing to prove. I will prove $\nu_2(s_{2^l (2k+1)}-1)=\nu_2(s_{2^l (2k-1)}-1)$ and a similar statement for $t$ to complete the induction. Note $s_{2^l (2k+1)}+t_{2^l (2k+1)}i=(s_{2^{l+1}} + t_{2^{l+1}}i)(s_{2^l (2k-1)}+t_{2^l (2k-1)}i)$, so we obtain $s_{2^l (2k+1)}-1=s_{2^{l+1}}s_{2^l (2k-1)} - t_{2^{l+1}}t_{2^l (2k-1)} - 1$. The conclusion follows by taking both sides mod $2^{l+3}$. We can do a similar thing for $t$. It remains to deal with $(1+2i)(-3+4i)^k=M-2i$. In this case, we can see $(-3+4i)^k + 1 =C(1-2i)$ for some integer $C$. So we want $-2(s_k+1)=t_k$. From the above, we can see $\nu_2(s_k+1)=1$, so $\nu_2(t_k)=2$, so $k$ must be odd. When $k=1$ this works and we obtain the solution $(11,2,3)$. Now, we note $(1+2i)^{2k+1} - (1+2i)^3 = (M-2i)-(11-2i)=M-11$, so $(-3+4i)^{k-1} -1 = \frac{M-11}{125} (-11-2i)$ so we are done again.

Here is a very simple solution. We will take 2 cases: --> $z$ is $odd$ * if $y \geq 2$, by taking $z=2z'+1$ we get: $$x^2 = 5.5^{2z'} -4^y \implies (x-2.5^{z'})(x+2.5^{z'}) = 5^{2z'} -4^y $$ with $x= 2.5^{z'} + x_1$ we obtain finaly: $$(5^{z'})^2 -4x_1.(5^{z'}) -x_1^2- 4^y =0$$ $$\Delta' = 5x_1^2 + 4^y \equiv 5 \mod (8)$$ for every $x_1$ $odd$ because the discriminant must be odd to get odd solutions to the equation. So no solution for this case. * $y=1$ The equation become by consider $x=5u+1$ and $5^{z'} = ku +1; k,u integers$ : $$x^2 - 1 = 5(5^{2z'} -1)\implies u(5u+2) = 5^{2z'} -1$$$$\implies (k^2-5)u= 2 -2k$$. Note that $|2-2k| \geq|k^2 -5)|$ for $-3\leq k \leq3$. So $u = -6-1;0;2$ are admissible and after verification solutions give: $u=0 \implies (x,y,z) = (1,1,1)$ $u=2 \implies (x,y,z) = (11,1,3)$ --> $z$ is $even$ Now we can write with $z=2z',$ $z'$ positive integer: $$4^y = (5^{z'} -x)(5^{z'} +x)$$ Note that one of the term in bracket is divisible by $2$ and the other by $4$. So by our conditions, we get $2 = 5^{z'} -x$ and plugging in the last equation gives: $$4.(5^{z'} -1) =4^y \implies 5^{z'} - 4^{y-1} = 1$$ By the help of Mihalescu's, the only solution is $$(z',y) = (1,2) \implies (x,y,z) = (3,2,2)$$ Note also that if it were allowed , $(x,y,z) = (2,0,1);(0,0,0);$ are also solutions. At the end, solutions are: $(x,y,z) = (1,1,1); (11,1,3); (3,2,2) / (2,0,1);(0,0,0).$