Using essentially a guess-and-check approach ...

There is a chance if both (x - 1) and (7x^2 - 3) are perfect squares.

squares: {0, 1, 4, 9, 16, ...}

from x-1, try x values: {1, 2, 5, 10, 17, ...}

7x^2 - 3 values: {4, 25, 172, 697, ...}

x=1 works to make LHS rational, but it does not solve the equation:

LHS=5 but RHS=3

x=2 works; and it does solve the equation:

LHS=RHS=4

**x=2 is a solution**, though perhaps not the only solution.

If the two radical terms had opposite signs, there might be an x value that would allow them to cancel each other out, but since there is a negative sign in front of each √, this cannot happen without complex numbers.

Running the program below shows no other x values where the two radicals are integers

Checking Wolfram Alpha shows that x=2 is the only real solution; and there are 4 complex solutions which I did not attempt to solve for.

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def issquare(n):

""" Input an integer, Returns True iff it is a perfect square. """

if round(n**0.5)**2 == n:

return True

else:

return False

xvalues = [i**2 + 1 for i in range(-100000,100000)]

xvalues = sorted(list(set(xvalues)))

for x in xvalues:

if issquare(7*x**2 - 3):

print(x, (x-1)**.5, (7*x**2 - 3)**.5)

print((x**2)*(7 -(7*x**2 - 3)**.5 - (x-1)**.5), x+2, '\n')