Find non-negative integers(mine)

[Masum]

Find all non-negative integers $(a,b)$ such that $7^a+11^b$ is a perfect square.

[Phlembac Adib Hasan]

Let $7^a+11^b=x^2$
It is very easy to see a is even and b is odd.(Use mod 3 and mod 4)
Now let $a=2k$
Then,$11^b=(x+7^k)(x-7^k)$
Now look that $x-7^k$ is either 1 or a power of 11.
If it is 1, then,
$7^a+11^b=7^a+2 \cdot 7^k+1$
$11^b-1=2 \cdot 7^k$
LHS is containing 5 as factor,but RHS not.It is not possible.
Now let's see if $x-7^k$ can be a power of 11.
Let it as an even power, means $x-7^k=11^{2m}$
Then $X+7^k=11^{2n+1}$ [Note that b is odd]
$11^{2m}+7^k=x=11^{2n+1}-7^k$
$2 \cdot 7^k=11^{2m} \cdot (11^{2n-2m+1}-1)$
Again RHS contains 5,but LHS not. Once again impossible.
Now it's time to check odd power:$x-7^k=11^{2p+1}$
Then $x+7^k=11^{2q}$
$11^{2p+1}+7^k=x=11^{2q}-7^k$
$2 \cdot 7^k=11^{2p+1}(11^{2q-2p-1}-1)$
Here RHS also contains 5, but LHS not. Absurd!
So, it has no solution.

Moderation Note: $L^AT_EX$ed on request.
*Mahi*

[afif mansib ch]

is my sol correct?
\[7^a+11^b=n^2\]
let a and b be even first.
\[7^2+11^2\neq n^2\]
so a,b,>2
from d.e.
\[7^{a/2}=2st \]
\[11^{b/2}=s^2-t^2\]
\[n=s^2+t^2\]
here 1 of s and t is even and s,t, are coprime
\[1\equiv 11^{b/2}\equiv 1-0(mod4)\]
so t=2r and \[7^{a/2}=4sr\]
bt any power of 7 can't be even.so it's a contradiction.
now let \[11^b=11^{2n+1}\]
\[11^{2n+1}=(n-7^{a/2})(n+7^{a/2})\]\
\[=11^k*11^t\]
let t>k.so\[2.7^{a/2}=11^t-11^k=11^x\]
which is also a contradiction.same way it can be proved for 7^(2n+1)
so such an integer doesn't exist.
so.... is it correct???

[Masum]

is my sol correct?
\[7^a+11^b=n^2\]
let a and b be even first.
\[7^2+11^2\neq n^2\]
so a,b,>2
from d.e.
\[7^{a/2}=2st \]
\[11^{b/2}=s^2-t^2\]
\[n=s^2+t^2\]
here 1 of s and t is even and s,t, are coprime
\[\ast 1\equiv 11^{b/2}\equiv 1-0(mod4)\]
so t=2r and \[7^{a/2}=4sr\]
bt any power of 7 can't be even.so it's a contradiction.
now let \[11^b=11^{2n+1}\]
\[11^{2n+1}=(n-7^{a/2})(n+7^{a/2})\]\
\[=11^k*11^t\]
let t>k.so\[2.7^{a/2}=11^t-11^k=11^x\]
which is also a contradiction.same way it can be proved for 7^(2n+1)
so such an integer doesn't exist.
so.... is it correct???

$\ast\to$ If $s$ even, then \[\ast 1\equiv 11^{b/2}\equiv 0-1(mod4)\]

[afif mansib ch]

tnnx via.bt if we do it like u showed we can again get to a contradiction.will it be correct then?

[Masum]

It is easier than you thought. If both of $a,b$ are even or both are odd, we reach that $n^2\equiv2\pmod4$, contradiction. Then one of $a,b$ is even. We have two cases $a=2k$ or $b=2y$. The two are almost similar. Let $a=2k$. Then \[(n+7^k)(n-7^k)=11^b\]
See both $n+7^k,n-7^k$ have only the prime factor $11$, so assume \[n+7^k=11^x,n-7^k=11^y\]
Subtract the second from the first \[2\cdot7^k=11^x-11^y\]
If $y>0$ then we must have $11|2\cdot7^k$, not possible. Thus, $y=0$. I think rest and the other case is trivial now.

[afif mansib ch]

ok i see.thanks via.

[vcrazy]

Plz somebody solve it..I'm very much weak in math...If anyone solve it and show me the math at length it'll be easier to get it well....thnx

Let n=2^9 * 3^4.
|) How many positive integer divisors of n^2 are less than n but don't divide n ?
||) What is the answer if n= 2^39 * 3^41?

[Tahmid Hasan]

Plz somebody solve it..I'm very much weak in math...If anyone solve it and show me the math at length it'll be easier to get it well....thnx

Let n=2^9 * 3^4.
|) How many positive integer divisors of n^2 are less than n but don't divide n ?
||) What is the answer if n= 2^39 * 3^41?

please post it as a new topic.

[Phlembac Adib Hasan]

Plz somebody solve it..I'm very much weak in math...If anyone solve it and show me the math at length it'll be easier to get it well....thnx

Let $n=2^9 . 3^4$
|) How many positive integer divisors of $n^2$ are less than $n$ but don't divide $n$ ?
||) What is the answer if $n= 2^{39}. 3^{41}$?

First define $ \pi (k) $ is the number of divisors of $k$.
Notice that $\pi (n^2)$ is itself an odd number. If $a$ divides $n^2$ and $a <n$, then $ \frac {n} {a} $ must be larger than $n$.So $n^2$ has equal divisors less than $n$ and greater than $n$.But if any number divides $n^2$ and it does not equal to $n$, then it must be smaller or larger than $n$.So any divisor of $n^2$ (without $n$ ) is in this category. Hence $n^2$ has $\frac {\pi (n^2)-1} {2} $ larger than $n$ or smaller than $n$.Again notice that if a number divides $n$, it must divide $n^2$. Since $n$ has $\pi (n) $ divisors in total, there are exactly $ \frac {\pi (n^2)-1} {2} - \pi (n) $ divisors of $n^2$ that are less than $n$ but does not divides $n$.

The next part is for beginners who does not know how to find $\pi (k) $.First do prime factorization of $k$.
Let it be like this: \[k=p_1^{q_1}.p_2^{q_2}...p_m^{q_m}\]
Here $p_1,p_2,...,p_m$ all are primes.Then you can find $\pi (k) $ like this:
\[ \pi (k)=(q_1+1)(q_2+1)...(q_m+1)\]
So our answers are $\frac {(2.9+1)(2.4+1)} {2} -(9+1)(4+1) $ and $\frac {(2.39+1)(2.41+1)} {2}-(39+1) (41+1) $,respectively.