2023/035) Prove that $4 . 29! + 5 ! \equiv 0 \pmod {31}$

We have as 31 prime  using Wilsons theorem

$30! + 1 \equiv 0 \pmod {31}$

or $30 * 29! + 1 \equiv 0 \pmod {31}$

as $30  \equiv - 1 \pmod {31}$

so we have  $-1 * 29! + 1 \equiv 0 \pmod {31}$

Or  $ 29! -1 \equiv 0 \pmod {31}$

Or $ 4 * 29! - 4 \equiv 0 \pmod {31}$

adding 124 which is multiple of 31 we get

$ 4 * 29! + 120 \equiv 0 \pmod {31}$

or $ 4 * 29! + 5! \equiv 0 \pmod {31}$

Proved 

2023/034) Given $a+b+c=3$ and $a^2+b^2+c^2 = 3 $ Solve for real a,b,c

We have 

$a+b+c=3\cdots(1)$

$a^2+b^2+c^2=3\cdots(2)$

Squaring (1) we get

$a^2+b^2 + c^2 + 2ab + 2bc + 2ca = 9$

Putting the value of $a^2+b^2+c^2$ from (2) we get

$ab+bc+ca = 3$

subracting above from (2) we get

$a^2+b^2+c^2 - ab - bc - ca = 0$

or $2a^2+2b^2+2c^2 - 2ab - 2bc - 2ca = 0$

or $(a^2-2ab + b^2) + (b^2-2bc+c^2) + (c^-2ca +a^2) = 0$

or $(a-b)^2 + (b-c)^2 + (c-a)^2= 0$

above is possible only if a=b= c

putting in (1) we get a = b = c = 1

2023/033) Show that if 5 does not divide n then 5$|n^4-1$

We have $n^4-1 = (n^2+1)(n^2-1)$ difference of 2 squares

 $= (n^2+ 1)(n+1)(n-1)$ difference of 2 squares

$= (n^2-4 + 5) (n+1)(n-1) as $n^2+ 1=n^2-4 + 5$ 

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

$= (n-2)(n+2)(n+1)(n-1)+ 5(n+1)(n-1)$

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

  $5(n+1))(n-1)$ is divisible by 5

$(n-2)(n-1)n(n+1)(n+2)$ being product of 5 consecutive numbers is divisible by 5

As n is not dvisible by 5 so $\frac{(n-2)(n-1)n(n+1)(n+2)}{n}$ is divisible by 5

Hence the sum is divisible by 5 and so the given expression

Note: this can be done faster using mod 5 arithmetic but I have chosen a different way.

Aditionaly this is direct from fermats little theorem 

2023/032) Given positive integers a,b,c and $a^3+b^3 = 2^n$ prove that $a=b$

Both a nd b have to be even or add as RHS is even as it has to be > 1.

Because if a is odd and b even or vice versa then LHS is odd.

Now let us consider 1st case that is a and bare odd

So we get

$a^3+b^3 = (a+b)(a^2-ab+b^2)$

a+b is even and $a^2-ab+b^2= a(a-b) + b^2$ is odd and it has to be 1

Or $a(a-b) + b^2=1$ 

The above is possble only if a-b = 0(else  or a = b and we get a = b = 1

Now of the even case 

Let $a=p.2^q$ and b= $m.2^r$ and without loss of generality let$ q >= r$ and p and m are odd.

So we get $p^32^{3q} + m^32^{3r} = 2^n$

Devideing both sides by $2^{3r}$ we get

$p^32^{3q-3r} + m^3 = 2^{n-3r}$

RHS is even as it is minimum 2

LHS is even if 3q=3r or q = r

Puttting it we get $p^3+q^3 = 2^{n-3r}$ 

As p and q are odd so both are same and hence $a= b = 1$ for odd and $a=b=2^x$ for even