We can solve the 2 equations and get $a=5$ and $b=2$ and hence $a^3-b^3=5^3-2^3=117$
but the above approach is longer as compared to the approach below
We have
$(a-b)^3 = a^3-3a^2b + 3ab^3 - b^3 = a^3-b^3-3ab(a-b)$
or $a^3-b^3= (a-b)^3 + 3ab(a-b)$
Putting the values we get $a^3-b^3=3^3 + 3 * 10 * 3 = 117$
In this process we have avoided solving of equations
$16p+1$ is perfect cube so let is be $x^3$
So $x^3-1 = 16p$
Or $(x-1)(x^2+x+1) = 16p$
Now $x^2+x+1$ is is odd so $16| x-1$
So $(x-1) = 16m$ and $x^2+x+1 = n$ where m and n are odd
So p = mn
As p is odd $m = 1, n = p$ or $m = p, n= 1$
n= 1 gives $x^2+x+1 = 1 $ or x = 0 which is not possible
so m= 1 and x= 17 giving n = 307 wihich is a prime
so $p = 307$
Le $P(a,b,c) = abc(a^3-b^3)(b^3-c^3)(c^3-a^3)$
Let us first find out $a^3 \mod 7$
We have $a^3 \mod 7 \in \{1,-1\}$ when $a \mod 7 \ne 0$
if $a \mod 7 = 0$ or $b \mod 7 = 0$ or $c \mod 7 \ne 0$ we have $7 | P(a,b,c)$
$a^3 \mod 7 \in \{1,-1\}$
$b^3 \mod 7 \in \{1,-1\}$
$c^3 \mod 7 \in \{1,-1\}$
each can take one of two values and there are 3 differences so one of them has has to be zero
hence $7 | P(a,b,c)$
Proved
we have
$x^2-xy+y^2 = 13$
we can complete square by addition of some thing from $y^2$ to $x^2-xy$
to get $(x^2-xy + \frac{y^2}{4}) + \frac{3y^2}{4} = 13$
or $(x-\frac{y}{2})^2 + \frac{3y^2}{4} = 13$
or multiplying by 4 we get $(2x-y)^2 + 3 y^2 = 52$
as it is sum of positive numbers we get $3y^2 <=52$ or $ y <=4$'
$y =1$ gives $(2x-y)^2 = 49$ or $2x-y=7$ as -7 shall give -ve x so x = 4, y= 1
$y=2$ gives $(2x-y)^2= 40$ not a square
$y=3$ gives $(2x-y)^2 = 25$ or $2x-y = 5$ giving x = 4 , y = 3 and $2x-y = -5$ gives -ve x
$y=4$ gives (2x-y)^2 = 4 $, $2x-y=2$ gives x = 3, y = 4
$2x-y=-2$ gives x = 1, y = 4
So we have solutions (4,1),(4,3),(1,4),(1,3)
We have $ 42 = 2 * 3 * 7$
We need to show that $3^p-2^p=1 \pmod {2}\cdots(1)$
$3^p-2^p=1 \pmod {3}\cdots(2)$
and $3^p-2^p=1 \pmod {7}\cdots(3)$
As $3^p$ is odd for any p and and $2^p$ us even we get
$3^p-2^p=1 \pmod {2}$
this is we have proved (1)
now p is odd so we have
$3^p$ is divisible by 3.
or $3^p \equiv 1 \pmod 3\cdots(4)$
now let us look as $2^p$ we have as p is odd so p = 2k+1
$2^p = 2^{2k+1} = 4^k * 2$
so $2^p \equiv 2 \pmod 3\cdots(5)$
From (4) and (5) we have
$3^p- 2^p \equiv 1 \pmod 3$
We have proved (2)
Now we need to check for mod 7
as p is greater that 4 and a prime p is of the form 6k+1 or 6k + 5
Let us take the 2 cases 6k+ 1 and 6k-+ 5
first 6k+ 1
as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem
$3^6 \equiv 1 \pmod 7$
so $3^{6k+1} = (3^6)^k .3 = 3 \equiv 3 \pmod 7$
Simlilarly
$2^6 \equiv 1 \pmod 7$
so $2^{6k+1} = (2^6)^k .3 = 2 \equiv 2 \pmod 7$
Or $3^p-2^p \equiv 1 \pmod 7$
So we have proved (3) for p = 6k+ 1
Next t 6k+ 4
as 7 is co-prime to 3 and 7 we have as per Fermats Little Theorem
$3^6 \equiv 1 \pmod 7$
so $3^{6k+5} = (3^6)^k 243 = 243 \equiv 3 \pmod 7$
Simlilarly
$2^6 \equiv 1 \pmod 7$
so $2^{6k+5} = (2^6)^k .32 = 32 \equiv 2 \pmod 7$
Or $3^p-2^p \equiv 221 \pmod 7$
Or $3^p-2^p \equiv 1 \pmod 7$
So we have proved (3) for p = 6k+ 5
We have proved (3) for both cases
as (1) (2) (3) all are proved hence Proved
We shall prove if by binomial expansion
we have $4^{2n}$
$= (5-1)^{2n}$
$ = \sum_{k=0}^{2n}{2n \choose k}5^{2n-2}(-1)^{k}$
$ = \sum_{k=0}^{2n-2}{2n \choose k}5^{2n-2}(-1)^k - {2n \choose 2n-1}*5 + 1$ separating last 2 terms
1st sum each term is divisible by $5^2$ that is 25
so we are left with
$4^{2n} \equiv - 10n +1 \pmod {25}$
or
$4^{2n}+10n \equiv 1 \pmod {25}$
we have $n^2 \equiv 0/1 \pmod 4$
So $m^2 + n^2 \equiv 0/1/2 \mod 4$
So any number of of the form 4k + 3 cannot be expressed as sum of 2 squares
There are infinitely many of them
Hence proved
