2017/008) Let P be a polynomial such that $P(x)=a_0+a_1x+?+a_nx^n$where $a_0,a_1,\cdots$? are non-negative integer. Given that P(1)=4 and P(5)=152 find P(6)

P(x) is a cubic polynomial as P(5) is 152 between $125(5^3)$ and $625(5^4)$
now coefficient of $x^3$ is 1
so $P(x) = x^3 + ax^2 + bx + c$
taking mod 5 we get c =2 so $P(X) = x^3 + ax^2+ bx + 2$ and so $a . 5^2 + 5b = 152-125-2 = 27$ giving a = 1, b= 2
sp $P(x) = x^3 +x ^2 + 2$ and it satisfies P(1) = 4
so P(6) = 254

2017/007) if $\alpha,\beta$ are the roots of $x^2-5x+1$ show that $\alpha^n+\beta^n$ is an integer and not divisible by 4 for any integer n

we have for n = 1 $\alpha+\beta = 5 $ which is integer not divsible by 4
and $\alpha\beta=1$
hence $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 5^2 - 2 = 23$ integer and not divisible by 4
hence it is true for n =1 and 2
let it be true for n = 1 to k for k >=2
now
$( \alpha^n+\beta^n)(\alpha+\beta) = \alpha^{n+1}+\beta^{n+1} + (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
or $\alpha^{n+1}+\beta^{n+1} = 5 (\alpha^n+\beta^n) - (\alpha^{n-1} + \beta^{n-1}$
or $\alpha^{n+1}+\beta^{n+1} =  4(\alpha^n+\beta^n) +  (\alpha^n+\beta^n)- (\alpha^{n-1} + \beta^{n-1} \alpha\beta$
Now let n be smallest integer so that $\alpha^{n}+\beta^{n}$ is divisible by 4
so $\alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2}$ is divsible by 4
so $\alpha^{n-3}+\beta^{n-3} = alpha^{n-1}+\beta^{n-1} - \alpha^{n-2}+\beta^{n-2} - 4alpha^{n-2}+\beta^{n-2}$
is divisible by 4 which is contadiction.
hence proved

2017/006) if $x= a + b $ $y = aw+bw^2$ and $z= aw^2+ bw$ then show that $x^3+y^3 + z^3 = 3(a^3+b^3)$ and $x^2+y^2 + z^2 = 6ab$

We have $x+y+z = a(1+w+w^2) + b(1+w+w^2) = 0$
so x^3 + y^3+ z^3 = 3xyz = 3(a+b)(aw+bw^2)(aw^2+bw) = 3(a+b)(a^2 w^3 + abw^2 + abw^4 + b^2w^3)$
= 3(a+b)(a^2 + ab(w^2+ w^4) + b^2) = 3(a+b)(a^2 + ab(w^2+w) + b^2)$
$ = 3(a+b)(a^2 - ab + b^) = 3(a^3+b^3)$
Further
$x^2 = a^2 + b^2 + 2ab$
$y^2 = a^2w^2 + b^2w + 2ab$
$z^2 = a^2w + b^2w^2 + 2ab$
hence $z^2+y^2+z ^2 = a^2(1+w+w^2) + b^2(1+w+w^2) + 6ab = 6ab$