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### Solve $2^x \cdot 3^y = 1 + 5^z$ in positive integers

If $z=6k+3,$ then $5^z+1$ is divisible by $7,$ so it cannot yield a solution. That finishes your solution, since it means that $x=y=1$ are the only solutions.

### solve in integers $d^4 \pm 1 = 2e^2, e = (pc^2 - 3d^2)/4.$

$\ \bullet d^4-1=2e^2$ This curve $v^2=2d^4-2$ has only two integral points, according to the magma online calculator as follows. IntegralQuarticPoints($[2,0,0,0,-2]$); It says that all integral ...

### solve in integers $d^4 \pm 1 = 2e^2, e = (pc^2 - 3d^2)/4.$

We have: $d^4-e^2=(d^2-e)(d^2+e)=(e-1)(e+1)$ 1): $e+1=d^2+e\rightarrow d=\pm 1$ $\Rightarrow pc^2-3=4e\rightarrow pc^2\equiv 3\bmod 4 \rightarrow c=\pm1 , p= 7, 11, 13, 17...$ because p is prime. ...

### Given $n$ integers $a_1$ to $a_n$ and an integer $K$, does there exist a solution which satisfies the following equation?

The coin problem, also called Frobenius problem, is probably what you are looking for: https://en.wikipedia.org/wiki/Coin_problem It is the problem of finding the largest $K$ that is not reachable, ...
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### number theory Diophantine equation real world example.

they used the Euclidean Algorithm to solve for $x=-7, y=4$ although they did not show it. here it is: First, $23=1\cdot 13+1\cdot 10, 13=1\cdot 10+1\cdot3, 10=3\cdot 3+1\cdot 1$, so we have: \begin{...
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### How to determine solvability of binary quadratic Diophantine equations of the form $x^2-axy+bx-y+c=0$?

the quadratic form part has square discriminant, therefore factors, and this continues .. Taking $z=x-ay,$ $$(1+ax)(1-ab -az) = 1 + a^2 c$$ So, you need to factor$1 + a^2 c$ where both factors ...
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### How to solve $p^2 - p + 1 = q^3$ over primes?

COMMENT.-Primes $p$ of the form $6x-1$ cannot be solution: in fact $$(6x-1)^2-(6x-1)+1=36x^2-18x+3=3(12x^2-6x+1)$$ so $q=3$ which is not compatible with $12x^2-6x+1=3^2$. It follows $p$ must be of the ...

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