Determine all pairs (x, y) of integers such that:
1+2^{x}+2^{2x+1}= y^{2}.
3 Likes
Please check my solution
Notice that if (x,y) is a solution so is (x,-y), so WLOG y ≥ 0.
Clearly y=0 doesn’t work
1+2^x+2^2x+1 ≥ 1,y>0
If x<-1, 1+2^x+2^2x+1 will not be an integer. If x=-1,1+2^x+2^2x+1=2≠y^2
If x=0 we get solutions (0,2),(0,-2)
Now we can say x>0.
1+2^x+2^2x+1=y^2 => 2^x+2^2x+1=y^2-1
=> 2^x(1+2^x+1)= (y+1)(y-1)
2 ∣ l.h.s , 2 ∣ (y-1)(y+1), both y+1,
y-1 have the same parity so both are even. gcd (y+1,y-1)=gcd(y+1,2)=2.
Case 1: v_2(y+1)=x-1, v_2(y-1)=1
v_2(y+1)=> y+1=2^x-1.m , y=2^x-1.m-1
s.t m ∈ N, m is odd.
By substituting in the original equation
1+2^x+2^2x+1=(2^x-1.m-1)^2
= 2^2x-2.m^2-2^x.m+1,
2^x+2^2x+1=2^2x-2.m^2-2^x.m
By dividing both sides by 2^x
1+2^x+1=2^x-2.m^2-m
=> 2^x-2(m^2-8)=m+1
If x=1, 1+2+8=11 ≠ y^2
So x=1 gives no solutions
If x=2, 1+4+32=37 ≠ y^2
So x=2 gives no solutions
So x>2.
=> 2^x-2(m^2-8)>2(m^2-8)
m+1>2(m^2-8), 17>2m^2-m => m<4
But m is odd so we test m=1,m=3
If m=1 , 2^x-2(m^2-8)=m+1
=> 2^x-2(-7)=2, no solutions.
If m=3 , 2x-2(m^2-8)=m+1
=> 2^x-2(1)=4, x=4
If x=4 , 1+2^4+2^9=529=23^2
So we get solutions (4,23),(4,-23).
Case 2: v_2(y-1)=x-1,v_2(y+1)=1
v_2(y-1)=x-1,y-1=2^x-1.m,
y = 2^x-1.m+1 , m ∈ N ,m is odd
1+2^x+2^2x+1=y^2=(2^x-1.m+1)^2
= 2^2x-2.m^2+2^x.m+1
=> 2^x+2^2x+1=2^2x-2.m^2+2^x.m
By dividing both sides by 2^x
1+2^x+1=2^x-2.m^2+m
1-m=2^x-2(m^2-8),1-m ≤ 0,2^x-2 ≥ 0
So m^2-8 ≤ 0, m^2 ≤ 8 < 9, m<3
But m is odd so m=1
But if m=1, 1-1=2^x-2(-7)
=> 0=2^x-2(-7), which is impossible.
So the only solutions for (x,y) is integers are (0,2),(0,-2),(4,23),(4,-23).