let z = x^2, then
z^2  6z + 1 = 7*2^y
add 7 to both sides:
z^2  6z + 8 = 7*(2^y + 1)
(z  4)(z  2) = 7*(2^y + 1)
Now the RHS is odd because it's the product of two odd factors, so z must also be odd. And since z is also a square, z = 1 mod 8.
The LHS mod 8 then is (1  4)(1  2) = 3 and so the RHS must also = 3 mod 8. Well, 7 = 1 mod 8 so (2^y + 1) = 3 mod 8 = 5 mod 8, and that means 2^y = 4 mod 8
But 2^y is a multiple of 8 whenever y >= 3, and so the only way 2^y = 4 mod 8 is if 2^y = 4 and y= 2
If y = 2, then the RHS = 7*(2^2 + 1) = 7*5 = 35
Since prime decompositions are unique, it must be that z2 and z4 are also 7 and 5, which requires that z = 9. Finally, since z = x^2 and x is positive, x = 3
The only solution in positive integers, then, is (x,y) = (3,2)

Posted by Paul
on 20230915 15:12:32 