if p and q are two natural numbers and p^q=81 then q^pq
Answers
Answer:
We're given two equations.
\begin{gathered}\\ \begin{equation}1981+p=q^2\end{equation}\\ \begin{equation}1981+q=p^2\end{equation}\end{gathered}
First we subtract (1) from (2).
\begin{gathered}\begin{aligned}&(2)-(1)\\ \\ \Longrightarrow\ \ &p^2-q^2=(1981+q)-(1981-p)\\ \\ \Longrightarrow\ \ &(p+q)(p-q)=1981+q-1981-p\\ \\ \Longrightarrow\ \ &(p+q)(p-q)=q-p\\ \\ \Longrightarrow\ \ &(p+q)(p-q)=-(p-q)\\ \\ \Longrightarrow\ \ &p+q=-1\end{aligned}\end{gathered}
⟹
⟹
⟹
⟹
⟹
(2)−(1)
p
2
−q
2
=(1981+q)−(1981−p)
(p+q)(p−q)=1981+q−1981−p
(p+q)(p−q)=q−p
(p+q)(p−q)=−(p−q)
p+q=−1
This doesn't hold true that p and q are distinct 'natural numbers' for the two equations taken simultaneously!
Because, since all the natural numbers are positive, then so will be the sum of at least two ones. If the sum of two natural numbers is negative, then one should be positive while the other should be negative.
\begin{gathered}\textsf{If}\ \ p\in\mathbb{N},\ \ p+q < 0\ \Longrightarrow\ q < -p\ \Longrightarrow\ q\not\in\mathbb{N}\\ \\ \textsf{If}\ \ q\in\mathbb{N},\ \ p+q < 0\ \Longrightarrow\ p < -q\ \Longrightarrow\ p\not\in\mathbb{N}\end{gathered}
If p∈N, p+q<0 ⟹ q<−p ⟹ q
∈N
If q∈N, p+q<0 ⟹ p<−q ⟹ p
∈N
But if p and q were not only distinct 'natural numbers', we could get the answer 10.
Let me show you.
Add (1) and (2).
\begin{gathered}p^2+q^2=1981+q+1981+p\\ \\ p^2+q^2=3962+p+q\\ \\ p^2+q^2=3962-1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\because\ p+q=-1]\\ \\ p^2+q^2=3961\end{gathered}
p
2
+q
2
=1981+q+1981+p
p
2
+q
2
=3962+p+q
p
2
+q
2
=3962−1 [∵ p+q=−1]
p
2
+q
2
=3961
Now, square the sum of p and q.
\begin{gathered}(p+q)^2=(-1)^2\\ \\ p^2+q^2+2pq=1\\ \\ 3961+2pq=1\\ \\ 2pq=-3960\\ \\ pq=-1980\end{gathered}
(p+q)
2
=(−1)
2
p
2
+q
2
+2pq=1
3961+2pq=1
2pq=−3960
pq=−1980
Hence answer will be,
1990+pq=1990-1980=\bold{10}1990+pq=1990−1980=10
Now we're going to find the unique values of p and q.
\begin{gathered}(p^2+q^2+2pq)-(4pq)=1-(4\cdot -1980)\\ \\ p^2+q^2-2pq=1-(-7920)\\ \\ (p-q)^2=7921\\ \\ p-q=89\end{gathered}
(p
2
+q
2
+2pq)−(4pq)=1−(4⋅−1980)
p
2
+q
2
−2pq=1−(−7920)
(p−q)
2
=7921
p−q=89
Finding p and q,
\begin{gathered}p=\dfrac{(p+q)+(p-q)}{2}=\dfrac{-1+89}{2}=\dfrac{88}{2}=\bold{44}\\ \\ \\ q=\dfrac{(p+q)-(p-q)}{2}=\dfrac{-1-89}{2}=\dfrac{-90}{2}=\bold{-45}\end{gathered}
p=
2
(p+q)+(p−q)
=
2
−1+89
=
2
88
=44
q=
2
(p+q)−(p−q)
=
2
−1−89
=
2
−90
=−45
Hence got!
The problems with distinct natural numbers p and q are that,
→ on taking p = 44 and q = 45, (1) is satisfied but (2) is not, because q should be -45 in the case of (2).
→ on taking p = -44 and q = -45, (2) is satisfied but (1) is not, because p should be 44 in the case of (1).