Write it as $(n-1)(n+1) = 5 \cdot 2^m$ $(n-1, n+1) = 2$ so if $2^t||n-1$ then $2||n+1$ and vice versa. So we have possibilities: $n-1=2$ $n+1=2$ $n-1=10$ $n+1=10$ From that we see that only $n=9$ works. $(n, m)=(9, 4)$

From(n-1)(n+1)=5. $2^{m}$ so (n-1)=$2^{k}$or(n+1)=$2^{k}$ Then $2^{k-1}=4$or$2^{k-1}=6$ So k=3,then (n=9,m=4)

we have 1 + 5.$2^{m}$ = $n^{2}$; 5.$2^{m}$ = $n^{2}$ - 1 = (n+1)(n-1); $2^{m}$ = $\frac{(n+1)(n-1)}{5}$ so n is odd and (n+1)(n-1) has only one prime 5 and the rest are all 2, wich means (n+1)(n-1) ends in 0, but 5 must divide one of them, so we have: n + 1 = 10 or n - 1 = 10, wich lead us to the unique solution (m,n) = (4,9)

tmbtw wrote: From(n-1)(n+1)=5. $2^{m}$ so (n-1)=$2^{k}$or(n+1)=$2^{k}$ Then $2^{k-1}=4$or$2^{k-1}=6$ So k=3,then (n=9,m=4) You should write k<m.

In the first case, $n+1=10 \implies (n = 9,m=4)$. In the second case, $n-1=5 \implies n = 6$, which yields no solutions.

I’m gonna Try to take a proper crack at this instead of plugging the solutions CLAIM: $m$ is an even integer and $n$ is an odd integer

Now Algebraic manipulations would lead us to find that $$5.2^m = (n-1)(n+1)$$Since this has to be true and $n$ is odd , Let $n=2k+1$ for some $k \in I^+$ Therefore $$5.2^m = (2k)(2k+2)$$$$\implies 5.2^m = 4(k)(k+1)$$Since $m$ is even we already have $4| m$ for all $m \in {2,4,6,8.....}$ $5=k$ or $5= k+1$ Therefore $$n+1 = 2.(k+1)$$$$\implies n+1 = 10$$ Therefore $ n = 9$ comes out as a solution and correspondingly $m= 2$ also comes out as solution CLAIM: There are no more solutions to the above.

This means that $5 \cdot 2^m = (n+1)(n-1)$. Now assume $n \neq 4, 9$. Then both $n-1$ and $n+1$ are something multiplied by a power of $2$. One must only be a power of two, and the other one must be $5$ multiplied by a power of $2$. Therefore, we must find positive integers $y, z$ satisfying $2^y+2=5 \cdot 2^z$. But the RHS is a multiple of $4$ when $z \geq 2$, so we only need to test $z = 0,1$. Out of these two, only $z = 1$ works, giving $y = 3$. Then the only solution is $n + 1 = 10, n - 1 = 8 \implies \boxed{(n,m) = (9, 4)}$.

Arne wrote: Find all positive integers $m$ and $n$ such that $1 + 5 \cdot 2^m = n^2$. We have $(n-1)(n+1)=5 \cdot 2^m$. We clearly have $\gcd(n-1,n+1)=1,2$. If the gcd is 1, then neither factor is divisible by 2, implying that $m=0$, which gives no solutions for $n$. Thus the gcd is 2. Evidently, this implies that both factors are divisible by 2, and exactly one is divisible by 4. If the factor divisible by 5 is also divisible by 4, then that factor is at least 20 while the other factor is 2, contradiction. Thus the v_2 valuation of the factor divisible by 5 is 1, so it is equal to 10. Checking, the only such solution is $(m,n)=(4,9)$, and no other solutions exist.

Arne wrote: Find all positive integers $m$ and $n$ such that $1 + 5 \cdot 2^m = n^2$.

Nice one! First, of all, $(n-1)(n+1)=5\cdot 2^m$, now, if $n$ is even, we have a contradiction. Now, if $n,$ is odd, then, both factors, are divisible by $2,$ and one of them is divisible by $4,$ it is a contradiction, if the factor, that is divisible by $5,$ is also a factor of $4.$ Now, we have that, we have that, we test all values less than $10,$ if we look at the powers of $2.$ Hence, we see, that $\boxed{(4,9)}.$

We first note that $m=0$ and $m=1$ yield $n^2=6$ and $n^2=11$ respectively, so there exists no integer solutions in these cases. Thus, $m\geq 2$ which gives us that $n>4$. Now, note that we have \[5 \cdot 2^m = n^2-1=(n-1)(n+1)\]and at most one of $n-1$ and $n+1$ is divisible by 5. Thus, we have two cases. Case 1 : $5 \mid n+1$. This means, that $2$ is the only prime dividing $n-1$ and so $n-1=2^a$. By our bound on $n$, this gives us that $a \geq 2$. But, notice that, \[n+1=2^a+2=2(2^{a-1}+1)\]which means $v_2(n+1)=1$ (since $a-1\geq 1$). Thus, we have \[m=v_2(5 \cdot 2^m) = v_2((n-1)(n+1)) = a+1\]Thus, $a=m-1$. Plugging this in we have $n=2^{m-1}+1$ and further, \[2^{m-1}(2^{m-1}+2)=2^m(2^{m-2}+1)\]which requires, $2^{m-2}+1=5$ which gives us that $m=4$ and in turn $n=9$. Case 2 : $5 \mid n-1$. This means $n+1=2^a$ and we have $a \geq 3$. Then, by a similar $v_2$ argument as before, we obtain that $a=m-1$. So, $n=2^{m-1}-1$. This gives, \[2^{m-1}(2^{m-1}-2)=2^n(2^{m-2}-1)\]which requires $2^{m-2}-1=5$ and thus, $2^{m-2}=6$ which clearly has no solutions. Thus, the only pair that works is $(4,9)$ which can easily be checked.