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Assuming that you mean √n + √(n + 4355):
First of all, we need n = x^2 for some integer x.
Next, we need n + 4355 = y^2 for some integer y
==> y^2 - x^2 = 4355
==> (y - x)(y + x) = 5 * 13 * 67.
Without loss of generality, we can take x > 0. Then, y - x < y + x.
Now, we simply consider all factorizations of 4355:
(i) y - x = 1 and y + x = 4355 ==> (x, y) = (2177, 2178)
(ii) y - x = 5 and y + x = 871 ==> (x, y) = (433, 438)
(iii) y - x = 13 and y + x = 335 ==> (x, y) = (161,174)
(iv) y - x = 65 and y + x = 67 ==> (x, y) = (1, 66).
Hence, there are four values of n (since n = x^2):
2177^2, 433^2, 161^2 and 1^2.
I hope this helps!