2

The number of ordered triples of nonnegative integers less than or equal to $2$ with at least one coordinate equal to $2$ (and hence having a maximum of $2$) is $3^3-2^3=27-8=19$ (all ordered triples, minus those with all coordinates less than $2$). Hence, since LCMs are all about taking the maximum of the exponents of each prime factor (and likewise, the ...


2

Hint $ $ You proved $\,6\mid P(k\!+\!1)-P(k),\,$ thus $\ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow\!\color{#c00}\Rightarrow 6\mid P(k).\,$ You used the $\,\color{#0a0}{{\rm direction}}\ \ 6\mid P(k\!+\!1)\color{#0a0}\Leftarrow 6\mid P(k)\,$ to inductively ascend its truth from $\,k\,$ to $\,k\!+\!1.\,$ Now use the reverse $\rm\color{#c00}{{\rm direction}\ (\...


2

These numbers are prime 1418575498609, 1418575498607, 1418575498603, 1418575498601, 1418575498597, 1418575498591, 1418575498589, 1418575498583, 1418575498579, 1418575498577, 1418575498573 which gives a gap of 42 between 1418575498571 and 1418575498613 They are 614101948*2310 - 1260 -[11,13,17,19,23,29,31,37,41,43,47]


1

To solve this question we need another condition. $$a=b$$ then solve value of a and b $$a+b=ab$$ replace a to b cause we know $$a=b$$ $$b+b=b*b$$ $$2b=b^2$$ $$b^2=2b$$ $$b^2-2b=0$$ $$b(b-2)=0$$ $$b=(0,2)$$ as we know $$a=b$$ meaning $$a=(0,2)$$ so, $$a+b=0+0$$ $$a+b=0$$ and $$a+b=2+2$$ $$a+b=4$$ so answer is a+b=(0,4)


1

I don't think that the Hasse principle is useful here. First, Hasse-Minkowski fails in general for cubic polynomials, e.g., for $3x^3+4y^3+5z^3$. For $x^3+y^3+z^3=n$ I don't see how to use it. Secondly, Ramanujan has used generating functions to obtain parametrised solutions for $x^3+y^3+z^3=1$ and $x^3+y^3+z^3=2$, and this direction appears to be more ...


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Write $x=18a$ and $y=18b$ where $a,b$ are coprime, so we have $$4a-9b = 6$$ Clearly $2\mid 9b$ so $2\mid b$ and thus $b=2d$. Similary we see that $3\mid a$ so $a=3c$. Now we have $c-3d=1$ and thus $c= 3d+1$ where $d$ is arbitrary. Now we get $$x= 54(3d+1)$$ and $$y= 36d$$


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Proof by induction. We can manually check a few numbers. $n = 3,7,11,15$ For each $n$ there as a corresponding $p = 3,7,11,3$ respectively. Suppose our proposition is true for all $n\equiv -1 \pmod 4$ and $n\le k$ Is our proposition true for the smallest $n >k$ such than $n\equiv -1 \pmod 4$? If $n$ is prime, we are done. If $n$ is composite, $n = ab$ ...


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