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9 votes

$a^2+a+3b$ and $b^2+b+3a$ both perfect square

You have the right general idea, but the following shows how to use perfect square lower and upper limits to determine the solution. First, due to symmetry, WLOG consider that $a \ge b$. Next, if ...
John Omielan's user avatar
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9 votes
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Does $23^n+1=2x^2$ have only two [positive integer] solutions?

Comments already contain enough information to craft an answer. Inspecting both sides mod $7$ we find that only possible remainder is $2$, i.e. $$ 23^n+1\equiv 2x^2\equiv 2\pmod 7. $$ Since $23^n\...
9 votes
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When can products of linear terms differ by a constant?

TL;DR: there are solutions for $n \leq 10$ and for $n=12$. Otherwise it's a well known open problem. See Wikipedia and here: http://euler.free.fr/eslp/TarryPrb.htm (archive). The question is ...
Bart Michels's user avatar
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9 votes
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Find prime numbers satisfying an equation

Let $m$ and $n$ be positive integers, and $p$ a prime number, such that $$p=\frac{m}{4}\sqrt{\frac{2n-m}{2n+m}}.$$ Then squaring and clearing denominators yields $$16p^2(2n+m)=m^2(2n-m).$$ Note that $...
Servaes's user avatar
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8 votes
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System of Pell's equations

Let $a$, $b$ and $k$ be positive integers such that \begin{eqnarray} 2k^2+1&=&a^2,\\ 6k^2+1&=&b^2. \end{eqnarray} Then clearly $k$ is coprime to $a$ and $b$ and $$a^2+(2k)^2=b^2,$$ so $...
Servaes's user avatar
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8 votes

find all rational numbers $x,y$ such that we have : $x^3+y^3=x^2+y^2$

This cubic has an isolated point at $(0,0)$; it is singular there. We can confirm this by checking that $f(x,y)=x^3+y^3-x^2-y^2$ is $0$ at that point and at its partial derivatives, $f(0,0)=f_x(0,0)=...
Merosity's user avatar
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7 votes
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Integral Solutions of $x^3+y^3+z^3+u^3=25$

Notice that $$ 25=27-2=3^3-2, \tag{1} $$ and $-2$ can be written as a sum of three cubes in infinitely many ways$^{(\dagger)}$, as $$ -2 = (-1-6c^3)^3+(-1+6c^3)^3+(6c^2)^3\quad(c\in\mathbb{Z}). \tag{2}...
Gonçalo's user avatar
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7 votes
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Extended Fermat's Last Theorem

We have the following non-trivial solution $$ 72^5=19^5+43^5+46^5+47^5+67^5. $$ Here $72$ is the smallest positive integer with this property. For more examples see here. In general, Euler's ...
Dietrich Burde's user avatar
7 votes

When can products of linear terms differ by a constant?

Assuming the respective $a_i$ and $b_i$ values are distinct, the values for $c$ you provide for degrees 2, 3, and 4 are minimal. The minimal values of $c$ for degrees $5$ and $6$ are given by the ...
mathmasterzach's user avatar
6 votes

Does equation $a^m+b^n =c^{m+n}$ have integral solutions?

For an arbitrary Pythagorean tripple $(a, b, c)$ with $a^2 + b^2 = c^2$, we have $(ca)^2 + (bc)^2 = c^4$, so $(ca, cb, c)$ forms a solution with $m = n = 2$. Another solution: If $a^2+b^3=c$, then ...
Sgg8's user avatar
  • 1,446
6 votes

Polynomial system of equations over integers

Taking the resultant over the complex numbers we obtain that either $(x,y)=(3,-4)$, or that $x$ is a root of an irreducible polynomial of degree $15$, having no rational root. So $x=3,y=-4$ is the ...
Dietrich Burde's user avatar
6 votes
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Polynomial system of equations over integers

If we subtract the equations we get to $$(y−1)^4−(x−1)^4=(y-x)(x+y-2)(x^2+y^2-2x-2y+2)=609=3\times 7\times 29$$ Since $3,7,29$ are primes then we have either: $y=x+3$ which solves to $(3,6)$ but not ...
zwim's user avatar
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6 votes
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Regarding $x^3 -3xy+y^3=2005$ in integer solution

Based on the posted question, I surmise that the official solution is poorly written. [E-1]: Given any integer $~r,~$ you have that the $~\pmod{9}~$ congruence class of $~r^3~$ will either be $~-1, ~0,...
user2661923's user avatar
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6 votes
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A system of six equations involving $a^3+b^3+c^3 = (c+1)^3$

There are many functions $f(x,y)$ such that the equation: $$x^3 + y^3 + (f(x,y))^3 = (f(x,y)+1)^3\tag1$$ has isolated solutions in integers, not forming cycles except in the trivial sense that ...
Adam Bailey's user avatar
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5 votes
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$n^6+18n$ is a perfect square if and only if $n=2$

If $\;n^6+18n=k^2\;$ where $\;n,k\in\Bbb N\;,\;$ then $n^3<k\;$ and $\;(k+n^3)(k-n^3)=18n\;,\;$ hence, $2n^3\!\!\!\!\!\underset{\overbrace{\text{because}\\\;n^3<k}}{<}\!\!\!k+n^3\leqslant18n\;...
Angelo's user avatar
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5 votes

Prime solutions to $p^2(p^3-1) = q(q+1)$

Suppose that $p$ and $q$ are prime numbers such that $$q(q+1)=p^2(p^3-1).$$ Clearly $q>p$ so $q$ divides $p^3-1$, but does not divide $p-1$. This implies $q\equiv1\pmod{3}$ and so $$p^2(p^3-1)\...
Servaes's user avatar
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5 votes
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Does the Diophantine equation $z_1^5 +z_2^5+z_3^5+z_4^5+z_5^5=\beta^5$ have a solution for every integer $\beta$?

This is a partial answer. There is a solution to $x_1 =22$, namely, $$22^5+42303^5+49013^5=51233^5+36563^5+3542^5$$ $$22+42303+49013=51233+36563+3542$$ I. Extended searches In 2017 I extended my ...
Duncan Moore's user avatar
5 votes
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Finding new solutions to $a^5+b^5+c^5+d^5+e^5=0$?

The answer to question 1 is YES. This is a consequence of the fact that, for all $n$, $n^5 \equiv 0 \pmod{11}$ iff $n \equiv 0 \pmod{11}$, otherwise $n^5 \equiv \pm1 \pmod{11}$. A somewhat more ...
Adam Bailey's user avatar
  • 4,207
5 votes
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$n^2-2^m = 1$, cant find the answer

You can rewrite that as $2^m = n^2 -1$ which is the same as $2^m = (n+1)\times(n-1)$ It immediately yields that both $n+1$ and $n-1$ are exact powers of 2. We can say that $\dfrac{n+1}{n-1}$ must be $...
Serge Ballesta's user avatar
5 votes
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On why solutions to $x^4+y^4+z^4 = 1$ come in pairs

Solutions were found with $d<10^{26}.$ I. New u $(u,v)=(\dfrac{1744}{495}, \dfrac{135}{1208})$ with rank $3$. Then $\pm \sqrt{D^2}$ gives the pair, $$372623278887^4+435210480720^4+369168502640^4=...
Tomita's user avatar
  • 2,366
5 votes

Solving diophantine equations: Finding integer solutions: $(x+1)^2+(x+2)^2+...+(x+2001)^2=y^2$

I think your problem does not have a solution. To see this, notice that $x^2 \equiv 1 \ \text{mod} \ 3$ whenever $x$ is not divisible by $3$. Further, $2001$ is divisible by $3$, thus, on the right ...
blomp's user avatar
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4 votes
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When is the norm of a ring $\Bbb Z[\sqrt{-p}]$ multiplicative?

For every nonsquare integer $d$ (positive or negative), the norm mapping ${\rm N} : \mathbf Z[\sqrt{d}] \to \mathbf Z$ where ${\rm N}(a+b\sqrt{d}) = a^2 - db^2$ (equivalently, ${\rm N}(\alpha) = \...
KCd's user avatar
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4 votes

Diophantine Equation $a^6+b^6+c^6=d^6$

This is related to a question of mine, but I think the answer there would be informative for this one. To find possible solutions to, $$a^6+b^6+c^6 = d^6\tag{i}$$ one way suggested by Piquito is to ...
Tito Piezas III's user avatar
4 votes

integral solutions of polynomials in two variables

The LHS $\,27x^4 - 256 y^3 = -\Delta\,$ where $\,\Delta = \prod_{i \ne j} (t_i-t_j)^2\,$ is the discriminant of the cubic $\,t^3 - 4 y t + x^2\,$ with roots $\,t_i\,$. Every such cubic with an integer ...
dxiv's user avatar
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4 votes
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Generalization of the Brahmagupta identity

There are more primitive forms that do this than just the principal ones. Meaning $\gcd(a,b,c)=1$ in $f(x,y) = ax^2 + bxy + c y^2 . $ In general, you ought to consider class number three or, at ...
Will Jagy's user avatar
  • 140k
4 votes
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Positive Integer solutions for $(x-y)^2 = \frac{4xy}{x+y-1}$

Just try and expand ? $$(x-y)^2(x+y-1)-4xy = (x-y)^2(x+y) - ((x-y)^2+4xy) = (x+y)(x^2-2xy+y^2-x-y) = 0.$$ Since you only want positive pair of solutions, we disregard $x+y = 0$ and then obtain: $$x_{1,...
dezdichado's user avatar
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4 votes
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Generalized Jacobi-Madden equation

There is a similar question I answered back in 2014. See the MSE post "Integer Solutions of $x^3+y^3+z^3 = (x+y+z)^3.\,$ In summary, I. 4th powers There are infinitely many solutions to, $$a^4+b^...
Tito Piezas III's user avatar
4 votes

How did Scott Chase find the only solution to eighth powers $x_1^8 + x_2^8 + x_3^8 + x_4^8 + \dots + x_8^8 = 1409^8$ back in 2000?

I don't know what S. Chase did, but a congruence relation that might have helped, assuming only $z$ and one left-hand side term $x$ are odd, is: Either $z+x\equiv 0\pmod{32}$ or $z-x\equiv 0\pmod{32}$...
Adam Bailey's user avatar
  • 4,207
4 votes
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$a^2+a+3b$ and $b^2+b+3a$ both perfect square

You made a good start. Here we set aside the trivial solution $a=b=0$. All you need to do next is check through a few possibilities for $m$ and $n$; most potential values for $m$ and $n$ can be ruled ...
John Bentin's user avatar
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