# Tag Info

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 ...
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 ... 1 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 $$y= 36d$$ 1 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\$ ...