we know that for a perfect square x $x^2 \equiv\, 0 or \, 4 \pmod 4$
and for $n \ge 4$ $n! + 3 \equiv\, 3 \pmod 4$
so for $n \ge 4 $ there is no solution so only n = 0 to 3 need to be checked
$0!+3 = 4 = 2^2$ is a perfect square
$1!+3 = 4 = 2^2$ is a perfect square
$2!+3 = 5 $ is not a perfect square
$3!+3 = 9 = 3^2$ is a perfect square
so n = 0 or 1 or 3
We are given
$2^{x+1} + y^2 = z^2 $
Hence $2^{x+1}= z^2 - y^2 = (z+y)(z-y)$
Because product of (z-y) and (z+y) is power of 2 hence each of them is a power of 2.
Let $z+y=2^a\cdots(1)$
$z-y= 2^b\cdots(2)$
Then $x+1 = a +b\cdots$
Adding (1) and (2) we get $2z=2^a+2^b=> z = 2^{b-1}(2^{a-b} + 1) $
And subtracting we get $2y=2^a+2^b => y = 2^{b-1}(2^{a-b} -1) $
y and z cannot be prime unless b =1 so we have b = 1
From above 2 we get $y = 2^{a-1} -1$ and $z=2^{a-1} + 1$
Now taking $y, 2^{a-1}, z$ we have $2^{a-1}-1,2^{a-1},2^{a-1} + 1$
3 consecutive integers and so one of them is divisible b 3. As $2^{a-1}$ is not divisible by 3 so either
$2^{a-1}-1$ or $2^{a-1}+1$ is. and it has to be 3 else it cannot be a prime. Now if $2^{a-1}+1$ is 3 then $2^{a-1} -1$ is 1 and not a prime.
so $2^{a-1} -1 = y = 3$ giving a = 2, z = y + 2 = 5 and putting it in the given equation z = 3
so solution set $(x=3,y=3,z = 5)$
We know that $343=7^3$
1st we know that $2^3-1 = 7$ or $2^3= 7 +1$
raise both sides to power 7 to get
$(2^3)^7= (7+1)^7 = \sum_{i=0}^7{7 \choose i}7^{i}1^{7-i}$
Except the 1st 2 terms each term is having a power of 7 greater than or equals to 2 so each term is divisible by $7^2=49$. the 1st term is 1 and the second term is $7 * 7$ or 49 so all except 1st term is divisible by 49 and 1st term is 1 so
$(2^3)^7= (1 + 49 q)$ for some q
or $2^{21}= 49q+ 1$ for some q
raise both sides to power 7 to get
$(2^{21})^7= (49q+1)^7 = \sum_{i=0}^7{7 \choose i}(49q){i}1^{7-i}$
Except the 1st 2 terms each term is having a power of 49 greater than or equals to 2 so each term is divisible by $7^4$ and hence $7^3=343$ The 1st term is 1 and the second term is $7 * 49q$ or 343q so all except 1st term is divisible by 49 and 1st term is 1 so
$2^{147}$ leaves a remainder `1 when divided by 343 or $2^{147}-1$ is divisible by 343
We have
$(1+a^{x_1}) (1+a^{x_2}) (1+a^{x_3})= 1 + a^{x_1} + a^{x_2} + a^{x_3} + a^{x_1+ x_2} + a^{x_2+ x_3 } + a^{x_3+ x_1 } + a^{x_1+x_2+ x_3}$
Applying AM GM inequality to $1, a^{x_1}, a^{x_2}, a^{x_3}, a^{x_1+ x_2}, a^{x_2+ x_3 }, a^{x_3+ x_1 }, a^{x_1+x_2+ x_3}$ we get
$\frac{(1+a^{x_1}+ a^{x_2} + a^{x_3} + a^{x_1+ x_2} + a^{x_2+ x_3 } + a^{x_3+ x_1 } + a^{x_1+x_2+ x_3}}{8}>=$
$\sqrt[8]{1* a^{x_1}* a^{x_2}* a^{x_3}* a^{x_1+ x_2}* a^{x_2+ x_3 }* a^{x_3+ x_1 }* a^{x_1+x_2+ x_3}}$
Or $\sqrt[8]{1* a^{4x_1+4x_2+4x_3}}$ Or 1 as $a^{4x_1+4x_2+4x_3} = a^0 = 1$
Or $\frac{(1+a^{x_1}+ a^{x_2} + a^{x_3} + a^{x_1+ x_2} + a^{x_2+ x_3 } + a^{x_3+ x_1 } + a^{x_1+x_2+ x_3}}{8}>= 1$
Or $(1+a^{x_1}+ a^{x_2} + a^{x_3} + a^{x_1+ x_2} + a^{x_2+ x_3 } + a^{x_3+ x_1 } + a^{x_1+x_2+ x_3}>= 8$
Or $(1+a^{x_1}) (1+a^{x_2}) (1+a^{x_3}) >=8$
Taking log to the base 2 we get the result
We are given
$x^3+ y^3 = 7\cdots(1)$
$x^2 + y^2 + x + y + xy = 4\cdots(2)$
Let us choose x+y = a and xy = b
then $1^{st}$ equation become
$x^3+y^3 = (x+y)^3 - 3xy(x+y) = 7$
or $a^3 - 3ab = 7\cdots(3)$
The $2^{nd}$ equation is
$x^2 + y^2 + xy + x + y = 4$
Or $(x+y)^2 - xy + x + y = 4$
putting $ x + y = a$ and $xy = b$ we get
$a^2 - b + a = 4$
or $a^2 + a - b = 4\cdots(4)$
multiplying (4) by 3a and subtracting (3) from it we get
$2a^3 + 3a^2 = 5$
or $2a^3 + 3a^2 -5= 0$
as a = 1 is a solution so we have
$2a^3 + 3a^2 - 5 = 2a^2(a-1) + 2a^2 + 3a^2 - 5 = 2a^2(a-1) + 5(a^2 - 1)$
$= 2a^2(a-1) + 5(a+1)(a-1) = 2a^2 + 5a + 5)(a-1) = 0$
so a = 1 or $2a^2 + 5a + 5 = 0$ this does not have any real solution
so a = 1
putting it in (3) we get $3ab = 1 - 7 = -6$
or b = -2
so we have x+y = 1 and xy = -2 giving 2 sets of solution (2,-1) or (-1,2)
so the solution set is
$(x, y) \in \{ (2,-1),(-1,2)\}$
