If $(n,m)$ is a solution of the equation,then obviously $n$ is odd,and $m \ge 2$ And we have $5\cdot 2^m=(n+1)(n-1)$.Because $(n+1,n-1)=2$.Then we have four cases: Case (a) $n-1=2\cdot 5,n+1=2^{m-1}$,No solutions. Case (b) $n-1=2^{m-1}\cdot 5,n+1=2$,No solutions. Case (c) $n-1=2,n+1=5 \cdot 2^{m-1}$,No solutions. Case (d) $n-1=2^{m-1}\cdot 5,n+1=2\cdot 5$,then $n=9,m=4$ Then the only solution is $(n,m)=(9,4)$.

$2^m|(n-1)(n+1)\to 2^{m-1}|n\pm 1\to n\ge 2^{m-1}-1$, because $(n-1,n+1)\le 2$. Therefore $5*2^m\ge 2^m(2^{m-2}-1)\to 2^{m-2}\le 6\to m\le 4$. Obviosly to chek, that $m=4,n=9$ is unique solution $m\le 4$.

5.2*m=(n+1)(n-1)/from euclid we know gcd(n+1;n-1)=2or1.n is odd so gcd(n+1;n-1)=2. Case 1:n-1=2/n=3/NO solution. Case 2:n+1=2/n=1/NO solution. Case 3:n-1=2.5/n=11/NO solution. Case 4:n+1=10/n=9/m=4 Lox sualdi.Basha dusduz eeeeeeeeeeeeeeeee. Lomsaniye. kishusuz es siz? bohohohohohohohohohohohohoo............../////////////...........

Another solution From $5*2^m = n^2-1$ we see that $n$ must be odd, thus $n=2n_1+1$ for some $n_1$. Thus $4n_1^2+4n_1 = 4(n_1^2+n_1) = 5*2^m$. This means that $m$ must be even or $m=2m_1$ making $4(n_1^2+n_1) = 5*4^{m_1}$. Thus $n_1^2+n_1 = n_1(n_1+1) = 5*4^{m_1-1}$. Obviously $n_1$ and $n_1+1$ are relatively prime so only one of them can contain a factor of 5. Here we split it into 2 cases $n_1$ is a multiple of 5, or $n_1 = 5n_2$ Then $5n_2(5n_2+1) = 5*4^{m_1-1}$. Thus $n_2(5n_2+1) = 4^{m_1-1}$. If $m_1 = 1$ then $m=2$ and we can check that this does not satisfy the requirements. Thus let $m_1 > 1$. This means that $n_2 = 4^a$ and $5n_2+1 = 5*4^a+1 = 4^b$ for some non-negative integers $a$ and $b$. But $5*4^a+1$ is not a multiple of 4 so the only possible power of 4 it could be is 1, but since $5*4^a > 0$ as $4^a > 0$ this is not possible. Thus the only case that could work is if $n_1$ is one less than a multiple of $5$, or $n_1 = 5n_2-1$. Then $(5n_2-1)5n_2 = 5*4^{m_1-1}$. Thus $(5n_2-1)n_2 = 4^{m_1-1}$. Now again $n_2 = 4^a$ and $5n_2-1 = 5*4^a-1 = 4^b$ for some $b$. If $a>b$ then since both $a$ and $b$ are integers we have that $a \ge b+1$ and that means $5*4^a - 4^b > 4*4^a-4^b = 4^{a+1} - 4^b \ge 16 - 1 = 15 > 1$ which doesn't work. Thus $b<a$ which implies $4^a(5-4^{b-a})=1$. Since both $4^a$ is positive we must have $5-4^{b-a}$ is positive and thus both are equal to 1. This means $a=0$ and $5*4^0-1 = 5-1 = 4 = 4^b$ meaning that $b=1$. Thus the only possible solution is $n_2 = 4^0 =1$ which makes $n_1 = 5*1-1=4$ meaning that $n=2*4+1 =9$ and plugging that in we get $m=4$ Thus the only solution is $(m,n) = (4,9)$

$ 5*2^m+1=n^2 $ from here we get that $ 5*2^m=(n-1)(n+1) $ and $ (n-1,n+1)=2,1 $ and the lefts side is even so it's 2. We have two possible cases: 1. If $ 5*2^k= n-1 $ and $ 2{}^m{}^-{}^k=n+1 $ and when we substract we get that $ 1+ 5*2^({}^k{}^-{}^1^) =2^({}^m{}^-{}^k{}^-{}^1^)$ so there aren't any solutions. 2. If $ 5*2^k= n+1 $ and $ 2{}^m{}^-{}^k=n-1 $ and when we substract we get that $ 1+ 2^({}^m{}^-{}^k{}^-{}^1^)= 5*2^({}^k{}^-{}^1^)$ so the right side is bigger or equal $5$ so $2^({}^m{}^-{}^k{}^-{}^1^)$ is even so the left side is odd and then so is the right side and we get that $k=1$ and then easily we get that that $m=4$ and $n=9$. So $n=9$ and $m=4$ is the only solution.

Lox lox lox

4dspace wrote: If $(n,m)$ is a solution of the equation,then obviously $n$ is odd,and $m \ge 2$ And we have $5\cdot 2^m=(n+1)(n-1)$.Because $(n+1,n-1)=2$.Then we have four cases: Case (a) $n-1=2\cdot 5,n+1=2^{m-1}$,No solutions. Case (b) $n-1=2^{m-1}\cdot 5,n+1=2$,No solutions. Case (c) $n-1=2,n+1=5 \cdot 2^{m-1}$,No solutions. Case (d) $n-1=2^{m-1}\cdot 5,n+1=2\cdot 5$,then $n=9,m=4$ Then the only solution is $(n,m)=(9,4)$. 4th step case

False written


$1+5 \cdot 2^m=n^2$ $\implies 5 \cdot 2^m=(n+1)(n-1)$ $\implies 10\cdot 2^k\cdot 2^{m-1-k}=(n+1)(n-1)$ $\forall$ $k\in \mathbb{N}_{0}$ $\bf Case$ $\bf 1:-$ $(n-1)=10\cdot 2^k$ , $(n+1)=2^{m-1-k}$ $\implies (m-1-k)\equiv 1\pmod{4}$ $2\equiv -1\pmod{3}$ $\implies 2^{4j+1}\equiv 2\pmod{3}$ $\forall$ $j\in \mathbb{N}_{0}$ $\implies 3|2^{m-1-k} - 2$ $\implies 3|(n+1)-2$ $\implies 3|(n-1)$ $\implies 3|10\cdot 2^k$ , which is a contradiction. $\bf Case$ $\bf 2:-$ $(n-1)=2^{m-1-k}$ , $(n+1)=10\cdot 2^k$ $\implies 2^{m-1-k}+2=10\cdot 2^k$ $\implies 2(2^{m-2-k}+1)=5\cdot 2^{k+1}$ $\implies k=0$ $\implies m=4$ , $n=9$

We are going to transform this equation into $5 \cdot 2^m=(n+1)(n-1)$ You will have few cases here: 1st: $n-1=5$ and $n+1=2^m$ - No natural solution. And you will have many other cases without natural solution but one case has natural solution: $n+1=10$ and $n-1=2^{m-1}\cdot 5$ $n=9$, $m=4$ is only natural solution.

ridgers wrote: Find all the natural numbers $m,n$ such that $1+5 \cdot 2^m=n^2$. We can see that $2^m \cdot 5=(n+1)(n-1)$ This implies that \[ 5\cdot 2^a \pm 2=2^b\]where $a+b=m$. So, Case 1: $5\cdot 2^a+2=2^b$, this implies that \[2^{b-1}=5\cdot 2^{a-1}+1\]This isn't possible, even though $a-1=0$. So no solutions here. Case 2: $5\cdot 2^a-2=2^b$, this implies that, \[2^{b-1}=5\cdot 2^{a-1} -1\]Here if $(a-1)=0$, then the following holds, where we get $b=3$ and $a=1$, So $m=a+b=3+1=4$ is the only solution, with $n=9$. It is similar to PRMO 2016 P1: https://artofproblemsolving.com/community/c4h1892025p12913370 Another Problem: If the problem asked us to find $(m,n)$ for \[5\cdot 2^m+(x)=n^2\]Where $x=\{2,3,5\}$ Then this will have no solution.. Because of Quadratic Residues!!!

$(n - 1 , n + 1) = (n - 1 , 2)=2 (or 1 : this in n=2 \implies no solution) $ $n - 1 < n + 1 \implies $ 1)$n - 1=2 and n + 1=2^m-1$ x 5 , no solution 2)$ n - 1=2^m-1 and $ n + 1= 2 x 5, $\implies m=4 , n=9 SOLVED $ $O.Y.SH.$

Case 1 : if $m=0$ then there in no solution Case 2 : if $m=1$ there is no solution Case 3 : So $m \ge 2$ LHS is odd so $n^2$ is odd, implies $n$ is odd Let, $n=2k+1$ $1+5\times2^m=4k^2+4k+1$ $5 \times 2^(m-2)=k(k+1)$ So $m>2,$ Sub-case 1 : If $k=5 \times 2^x$ for $0<x<m-2$ then $k+1=2^m-2-x,$ both of which are odd so they can't be consecutive Sub-case 3 : If $k+1=5 \times 2^x$ then $k=2^{m-2-x}$ which is not possible similarly So, $x=0,$ plugging this in both sub-cases gives one solution(from sub-case 2) $k=4, k+1=5$ so, $m=4, n=9$

ridgers wrote: Find all the natural numbers $m,n$ such that $1+5 \cdot 2^m=n^2$. We have $5\cdot 2^m=(n-1)(n+1)$ $\Rightarrow n-1=2^k$, or $n+1=2^k$ If $n-1=2^k$ $\Rightarrow n+1=2^k+2$ But $5|2^k+2$, $2^k+2=2^t\cdot 5$ $\Rightarrow t=1, k=3 \Rightarrow n=9,m=4$ If $n+1=2^k$ $\Rightarrow n-1=2^k-2$ But $5|2^k-2$, $2^k-2=2^t\cdot 5$ $\Rightarrow t=1,2^k=12$ $(\Rightarrow\Leftarrow)$ $(m,n)=(4,9)$ their is a unique solution

$1+5.2^m= n^2 \\ \Leftrightarrow 5.2^m= (n-1)(n+1) \\ \Leftrightarrow 5.{2^{m-2}}=\frac{(n-1)}{2}\frac{(n+1)}{2} \\ GCD (\frac{(n-1)}{2};\frac{(n+1)}{2})=1 \\ 1) \ \frac{(n-1)}{2}=1 \Leftrightarrow n=3 \ But \ n=3 \ is \ not \ a \ solution. \\ 2) \ \frac{(n-1)}{2}=5 \Leftrightarrow n=11 \ But \ n=11 \ is\ not \ a \ solution. \\ 3) \ \frac{(n+1)}{2}=5 \Leftrightarrow n=9 \ , \ n=9 \ is \ solution \ with \ m=4. \\ 4) \ \frac{(n+1)}{2}=1 \Leftrightarrow n=0 \ But \ n=0 \ isn't \ natural. \\ At \ the \ end \ (n;m)= (9;4) \ is \ the \ only \ solution \ to \ the \ problem. $

We have, $$1+5\cdot 2^m=n^2 \Rightarrow 5\cdot 2^m=n^2-1=(n-1)(n+1)$$Since $n$ is odd and $n^2 \ge 1+5\cdot 2= 11 \Rightarrow n\ge4$, there exist a positive integer $l$ s.t. $n=2l+1$. We have $(n-1)=2l$ and $(n+1)=2(l+1)$, now we substitute it to the previous equation. $$5\cdot 2^m=(n-1)(n+1)=4l(l+1) \Rightarrow 5\cdot 2^{m-2}=l(l+1)$$It's obvious that $gcd(l,l+1)=1$, since $l+1\ge 1+1=2$ there are 3 cases. Case 1. $(l,l+1)=(1,5\cdot 2^{m-2})$ Since $l=1$, $5\cdot 2^{m-2}=l+1=2$. There are no positive integer solution in this case. Case 2. $(l,l+1)=(5,2^{m-2})$ Since $l=5$, $2^{m-2}=l+1=6$. There are no positive integer solution in this case. Case 3. $(l,l+1)=(2^{m-2},5)$ Since $l+1=5$, $2^{m-2}=l=4 \Rightarrow m=4$. Because $l=4$, therefore $n=2l+1=9$. The only solution in this case is $(m,n)=(4,9)$. Having all cases exhausted, the only solution is $(m,n)=(4,9)$.

LHS is odd so $n^2 \equiv 1 \pmod 4$ which means that $m \ge 2$, now we also have $n \equiv \pm 1 \pmod 5$. Now let $n=2k+1$ then we have that $k(k+1)=5 \cdot 2^{m-2}$ and as $\gcd(k, k+1)=1$ we have that either $k=1$ (which trivially) fails or one of them is $5$ and the other a power of two, clearly if $k=5$ the. $k+1=6$ is not power of two so $k+1=5$ meaning $k=4$ which is power of two, this also gives $m=4$ therefore $(m, n)=(4, 9)$ is the only pair that works thus we are done .

(m,n)=(4,9) by moving 1, factoring right side and doing some casework