$P$ is an integer . $ Q$ is an odd positive integer.
If $P^2 + 17^Q = 10^8$ , what is the number of values that $P$ can take?
Given $P^2 + 17^Q = 10^8 $, what is the number of values that $P$ can take?
1
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number-theory
elementary-number-theory
diophantine-equations
1 Answers
5
There is no such $P$, thus the answer is $0$.
The reason is because $x^2 \equiv 0,1 \pmod {4}$ for all $x \in \mathbb{Z}$.
Assume there exists such integer $P,Q$, that $P^2+17^{Q}=10^{8}$, Tten we have that$$P^2=10^{8}-(16+1)^{Q} \equiv 0-1 \equiv 3 \pmod {4}$$ So $P^2 \equiv 3 \pmod {4}$. Contradiction to the statement above.
You can see this is true no matter the value of $Q$, so the condition that $Q$ is odd is unnecessary.
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0But if there was a - instead of a + then it would at least not be impossible for that reason? – 2017-02-25
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0@mathreadler I don't know. There would be a finite amount of $P$ still, I can guarantee that much. – 2017-02-25