2023/024) show that $(x+1)^7−x^7−1$ is divisible by $x(x+1)(x^2+x+1)$

First we note that $gcd(x,x+1) = 1$

$gcd(x, x^2 + x + 1) = GCD(x, x(x+1)) = 1$

and $gcd(x+1, x^2 + x + 1) = GCD(x+1, x(x+1)) = 1$

So $x,x+1,x^2+x+1$ are pairwise co-prime and to prove that the given expression is divisible by  $x(x+1)(x^2+x+1)$  we need to show that it is divisible by $x$, $x+1$ and $x^2+x+1$

Now Let $P(x) = (x+1)^7−x^7−1$

To show that it is dvisible by x we need to show P(0) = 0

$P(0) = 1^7 - 0^7 -1 = 0$  so it is divisibe by x

To show that it is dvisible by x+1 we need to show P(-1) = 0

$P(-1) = (-1+1)^7 - (-1)7 -1 = 0$  so it is divisibe by x+1

Now we need to show that it is divisible by $x^2+x+1$

That is if $x^2+x+1=0$ then $P(x) = 0$

Because 

$x^2+x+1=0\cdots(1)$ we get

$x+1 = - x^2\cdots(2)$

And multiplying by $x-1$ we have $x^3-1 = 0$ 

Or $x^3 = 1\cdots(3)$

Now $P(x) = (x+1)^7−x^7−1$

$= (-x^2)^7 - x^7 -1$ using (1)

$=x^{14} - x^7 -1$

$= -(x^3)^4 * x^2 - (x^3)^2 * x - 1$ using (2)

$= -x^2 - x - 1$

$= - (x^2+x+1)=0$ using (1)

So P(x) =0 if $x^2+x+1=0$

Hence $x^2+x+1$ is a factor

As $x,x+1,x^2+x+1$ are factors and are pairwise co-prime so   $(x+1)^7−x^7−1$  is divisible by $x(x+1)(x^2+x+1)$ 

2023/023) How do you prove that $n^4−1$ is divisible by 5 if n is not a multiple of 5?

We have $n^4 -1 = (n^2+1)(n^2–1) = (n^2–4 + 5)(n^2–1) = (n^2–4)(n^2–1) + 5(n^2–1)$ 

$= (n+2)(n-2)(n+1)(n-1) + 5(n^2–1)$

now the 2nd term is multiple of 5 and 1st term is product of 4 numbers which along with are 5 consecutive numbers. so one of them has to be divisible by 5. but as n is not divisible by 5 so one of the 4 other numbers is a multiple of 5 and hence the product. as given number is sum of 2 numbers each multiple of 5 and hence the given expression.

2023/022) Show that $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$ is an integer for any integer n

 We have $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$

$=\frac{n^7-n +n }{7} + \frac{n^{13}-n + n}{13} + \frac{71n}{91}$ 

$=\frac{n^7-n} {7} + \frac{n^{13}-n}{13} + \frac{n} {7}+ \frac{n}{13} + \frac{71n}{91}$ 

$=\frac{n(n^6-1)}{7} + \frac{n(n^{12}-n) }{13} + n$

If n is divisble by 7 1st term is integer and if n is not divisibl by 7 $n^6-1$ is divsible by 7 as per Fermats Little Theorem is 1st term is integer similarly the 2nd term   and 3rd term is integer for any integer n and hence the given expression  

2023/021) Find the sum of the series $1+5+10+16+23\cdots$

 Let us have a look at the $t^{th}$ term. 

Let us look at difference for a couple of terms

$t_2 = t_1 = 5- 1 = 4$

$t_3-t_2 = 10 -5 = 5$

$t_4-t_3 = 16 -10 = 6$

From the above we see that $t_k = t_{k-1} + k + 2$

From this let us find $t_n$  in terms of n

We have $t_n = 1 + \sum_{k=2}^{n} (k + 2)$

$= 1 + \frac{n(n+1)}{2} - 1 + 2(n-1)$

$= \frac{n^2}{2} + \frac{5n }{2} - 2$

So sum $S_n = \sum_{k=1}^n (\frac{k^2}{2} + \frac{5k }{2} - 2)$

$= \sum_{k=1}^n (\frac{k^2}{2}) + \sum_{k=1}^n\frac{5k }{2} - \sum_{k=1}^n2$

$=\frac{1}{2}(\frac{n(n+1)(2n+1)}{6}  + \frac{5}{2} \frac{n(n+1)}{2} - 2n $

$= \frac{2n^3 + 3n^2 +n }{12} + \frac{5n^2 +5n}{4} - 2n$

$= \frac{2n^3 + 3n^2 + n + 15n^2 + 15n -24n}{12}$

$= \frac{2n^3 + 18n^2 -8n}{12}$

$= \frac{n^3 + 9n^2 -4n}{6}$

 

 

 

2023/020) Find the minumum of $\sqrt{(x^2 + 4x + 13)} + \sqrt{(x^2 -8x + 41)}$

 We have 

 $\sqrt{(x^2 + 4x + 13)}$ Completing the square we get

$= \sqrt{x^2+ 4x+ 4 + 9} = \sqrt{(x+2)^2 + 3^2}$

The above in a plane is the distance from $(-2,-3)$ to $(x,0)$

 $\sqrt{(x^2 - 8x + 41)}$ Completing the square we get

$= \sqrt{x^2 - 8x+ 16 + 25} = \sqrt{(x-4)^2 + 5^2}$

The above in a plane is the distance from $(4,-5)$ to $(x,0)$ also from (4,5) to $(x,0)$ and because of symmetry as the line has to be to $(x,0)$ we can choose $(4,-5)$

The sum of the 2 expression is the distance from $(-2,3)$ to $(x,0)$ and from $(x,0)$ to $(4,5)$ or $(4,-5)$ as the 2 distances are same. it is better to chose a point on the other side of x axis so (4,5).

The distance ins minimum if  $(-2,3),(x,0)$ and $(4,5)$ are in same line and so minimum distance is distance from $(-2,3)$ to $(4,5)$ $= \sqrt{(4+2)^2 + (5+3)^2} = 10$

2023/019) Let a,b,c,d be any four real numbers but not all equal to zero. Prove that the roots of the polynomial $x^6 + a x^ 3 + bx^2 + c x + d = 0$ cannot all be real.

Because it is a degree 6 polynomial so  there are six roots let them be $x_1,x_2,x_3,x_4,x_5,x_6$ 

Because coefficient of  $x^5$  is zero so sum of roots is zero

So $\sum_{i=1}^{6}x_i = 0\cdots(1)$

Because coefficient of  $x^45$  is zero so double sum of roots is zero

So $\sum_{i=1}^{6}\sum_{j=1. j\ne i}^{6}x_i x_j = 0\cdots(2)$

Hence from (1) and (2) we have $\sum_{i=1}^{6}x_i^2 = 0$

If roots are real  all $x_i$ are zero which is impossible if a,b,c,d are non zero

Hence there is a contradiction and all roots cannot be real 

2023/018) Can the sum of three squares of odd numbers be a perfect square

An odd number is of the form 2n+ 1, so let 3 odd numbers be 2a+1, 2b+1,2c+ 1

Now $(2a+1)^2 = 4a^2 + 4a + 1 = 4(a^2+ a) + 1$

$(2b+1)^2 = 4b^2 + 4b + 1 = 4(b^2+ b) + 1$

$(2c+1)^2 = 4c^2 + 4c + 1 = 4(c^2+ c) + 1$

We observe that square of an odd is of the from 4n + 1

Now $(2a+1)^2 + (2b+1)^2 + (2c + 1)^4 = 4(a^2 + a + b^2 + b + c^2 + c) + 3$

Above is an odd number and of the form 4n + 3 so cannot be a perfect square 

So the answer is No