Are Transitions in a Hydrogen Atom Unique So there was a question on a past exam paper of a test I have recently taken and despite the test being over I feel the need to know the answer. I am a physics major and the test was a generic test on problems similar to the GRE in the States but with more involved questions.
The question asked;
Are there any two transitions in a hydrogen atom which will emit the same wavelength of electromagnetic radiation. (Ignore any degeneracy in l and s etc.)
My Thoughts;
Now I know that for hydrogen energy levels;
$$ E_n = \frac{c}{n^2} $$
Where c is some constant and n is the principle quantum number.
Thus the energy for a transition from level $n$ to level $m$ $(n>m)$ is given by
$$ E_{n-m} = c\bigg(\frac{1}{n^2}-\frac{1}{m^2}\bigg)$$
So the question may be reduced to finding any integer solutions $(m, n, p, q \in  N)$ to
$$ \bigg(\frac{1}{n^2}-\frac{1}{m^2}\bigg) = \bigg(\frac{1}{p^2}-\frac{1}{q^2}\bigg) $$
My intuition for the answer has oscillated. I am currently of the opinion it's not possible but could be wrong. I have attempted this a few ways. One way was to multiply across by $n^2m^2p^2q^2$ and gather on one side.
$$ m^2 p^2 q^2 - n^2 p^2 q^2 - n^2 m^2 q^2 + n^2 m^2 p^2 $$
Then I tried to prove that this couldn't be equal to 0. As it was always positive or negative. One I also notes was that if we say for the trasition to be uniques. $ n > p > m > n $. If this was not true then one trasition would be greater than the other. If this question can be solved a differnent way Id like to know but I'd also be interested in how to prove the above identity regardless.
Apologies if Ive missed something obvious.
 A: I wrote a little Python program to find them:
#!/usr/bin/env python3

max_n = 200
for n in range(1, max_n):
    for m in range(n+1, max_n):
        for p in range(1, n):
            for q in range(p+1, m):
                if m*m*p*p*q*q +n*n*m*m*p*p == n*n*p*p*q*q + n*n*m*m*q*q:
                    print(f"Coincident spectral line for m={m}, n={n}, p={p}, q={q}")

Definitely not unique!
Coincident spectral line for m=90, n=6, p=5, q=9
Coincident spectral line for m=35, n=7, p=5, q=7
Coincident spectral line for m=56, n=8, p=7, q=14
Coincident spectral line for m=72, n=8, p=6, q=9
Coincident spectral line for m=72, n=9, p=6, q=8
A: Above equation shown below:
$m^2 p^2 q^2 - n^2 p^2 q^2 - n^2 m^2 q^2 + n^2 m^2 p^2=0$
Above is equivalent to:
$ (m^2 p^2 q^2) +(n^2 m^2 p^2)
= (n^2 p^2 q^2) + (n^2 m^2 q^2)$
Or
$(mpq)^2+(nmp)^2=(npq)^2+(nmq)^2$ -------(A)
The above can be replaced by:
$(ac+bd)^2+(ad-bc)^2=(ac-bd)^2+(ad+bc)^2$ -------(B)
For equation (A) solution given by (User 14717) is (m,n,p,q)=(90,6,5,9)
Hence the solution for equation (B) is:  (a,b,c,d)= (2,1,1080,1890)
But equation (B) has the condition:
$c^2(12a^2+b^2)=d^2(a^2+12b^2)$  -------(C)
Hence any new numerical solution to equation (C) will give new solution to Equation (A)
