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Find all pairs $(m,n)$ such that the quotient $q$ and reminder $r$ of $\frac{m^2+n^2}{m+n}$ satisfies $q^2+r=17$

Clearly, $1\leq q \leq 4$. We can consider the 4 possible values of $q$ in turn. If $q=1$, then $r=16$, and we have $m^2+n^2 = 1(m+n)+16$, or $(2m-1)^2 + (2n-1)^2 = 66$, which has no solutions $m, n \...
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Is there a "magic" binary operation from which addition and multiplication can be derived?

A solution close to what you proposed. With 3 registers $r, s, t$, the operation $r + st$ is sufficient for addition and multiplication. $r = x, s = 1, t = y$ results in $x + y$. $r = 0, s = x, t = y$ ...
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Diophantine equation $\frac{x^4 + y^4}{z^4 + t^4} = u^2$

Here is a table with small solutions ($x \le y \le 7300$, ordered by $y$) for the equation $$x^4+y^4 = u^2(z^4+t^4);$$ it is divided into $3$ parts: $u=1$; $u=41$; other values for $u$. \begin{array}...
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1 vote

Is there a "magic" binary operation from which addition and multiplication can be derived?

OISC or "One Instruction Set Computer" is not Possible, because it means all Programs are just the single OPCODE repeated, which means all Programs are Exactly the Same ! You must have ...
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Wieferich Criterion proof verification

This could work: Theorem: If: $$x^p=y^p+z^p$$ $$p \not | xyz$$ $$gcd(x,y,z) =1$$ Then: $$2^{p-1} \equiv 1 \pmod{p^2}$$ Proof Let: $$az \equiv x-py \pmod{p^3}$$ $$by \equiv x-pz \pmod{p^3}$$ $\implies$ ...
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2 votes

Is it possible to generalize this equation more?

This is very nice. It relies to the Tarry-Escott problem, on which there is a vast literature going back more than 100 years. I do not know how much you know of this. I suggest you consult Dickson, ...
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1 vote

A cryptogram Diophantine $\overline{ABCDE}=4 \ \overline{AB}^2+\overline{CDE}^2$

Since $100 \le \overline{CDE} < \sqrt{\overline{ABCDE}} \le \sqrt{99999} \approx 316.226$, we must have $C \in \lbrace 1, 2, 3 \rbrace$. Doing arithmetic modulo 10, we have $E = 4B^2 + E^2$. This ...
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3 votes
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A cryptogram Diophantine $\overline{ABCDE}=4 \ \overline{AB}^2+\overline{CDE}^2$

We convert the given equation to the following form. $$1000(AB)+(CDE)=4(AB)^2+(CDE)^2. $$ Now consider the following quadratic equation. $$(CDE)^2-(CDE)+4(AB)^2-1000(AB)=0$$ The two roots of this ...
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Solve $\log_2(x+1)=⌊\log_2(x)⌋+1$ in positive integers

In $$\log_2(x+1)=⌊\log_2(x)⌋+1\tag 1$$ the right side is an integer, thus $x+1$ must be an integer power of $2$. Hence $x+1=2^n$ with $n\in\Bbb Z$. In addition, we have that $x$ must be positive (...
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Find all positive integer solutions verifying two conditions

Consider $x=2^y+2n-1$, where $2n-1<2^{y+1}-2^{y}$, then $(x+1)^2$ is a multiple of $2^y$ only when $n=2^{y-1}$. If $x$ is even, then no solutions can exist because it leads to the conclusion that ...
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1 vote

Continuing with Brahmagupta-Fibonacci identity...

\begin{cases} ax+by=24 \\ ay-bx=7 \end{cases} Solution with Brahmagupta-Fibonacci Identity: \begin{align}&(ax+by)^2+(ay-bx)^2=(a^2+b^2)(x^2+y^2)=24^2+7^2=625. \\&(a^2+b^2, x^2+y^2)=(5, 125), (...
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Show that $x^2 - 3y^2 = n$ either has no solutions or infinitely many solutions

There are an infinite number of solutions to $x^2-3y^2=1$, which you can generate from the base solution $2^2-3\cdot 1^2 = 1$ as follows: Given $(x,y) \in \mathbb{Z}^2$ such that $x^2-3y^2 = 1$, [ x^2-...
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Number of real and integer solutions of a multivariate polynomial

There is no such bound. Given an equation $X$ over $\mathbf{Q}$ with infinitely many rational points, you can always scale the coefficients so that any finite subset of the rational points all become ...
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solve in positive integers $\frac{1}a + \frac{1}b + \frac{1}c = \frac{4}5$

The crucial observation is: if $a$ is too big, then $\frac{1}{a}$ is too small (and analogously for $b$ and $c$). Hence, the potential solutions involve natural numbers that have upper bounds. It's ...
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5 votes
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solve in positive integers $\frac{1}a + \frac{1}b + \frac{1}c = \frac{4}5$

One way to do this is casework on the smallest denominator. Assume without loss of generality that $a\leq b\leq c$. Then $$\frac 45=\frac 1a+\frac1b+\frac1c\leq \frac 3a\implies a\leq \frac{15}4,$$ so ...
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2 votes
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Is there a better way to finish this number theory/algebra problem other than trial and error?

Yes, there is a way to solve it using a Pell equation. We want to find $N$ and $K$ such that $$2N^2 = K^2 + K.$$ Multiply by $4$ on both sides and add one to get $$8N^2 + 1 = 4K^2 + 4K + 1.$$ You can ...
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2 votes

Is there a better way to finish this number theory/algebra problem other than trial and error?

Write it as $8N^2+1=4K^2+4K+1=(2K+1)^2=2(2N)^2+1$ and you have the standard Pell equation. Alice represents the solution $2K+1=17, 2N=12.$ Bob represents the solution $2K+1=99, 2N=70$. The next ...
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solve in integers $d^4 \pm 1 = 2e^2, e = (pc^2 - 3d^2)/4.$

In your argument, when you get $u^2=d^2\pm1$, you have two integer squares differing by 1, so $u$ and $d$ can only be 0 and 1, giving only the solution $d=1$, $e=0$. The other solutions arise from $d^...
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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.
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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 ...
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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. ...
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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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1 vote
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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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