4=2^2

9=3^2

25=5^2

36=3^2 +3^2 + 3^2 +3^2

49=7^2

64=16*(2^2) or 4*4^2 or 3*4^2 + 4*2^2 or ....

A number which is equal to the sum of the squares of its prime factors with multiplicity:
- $16=2^2+2^2+2^2+2^2$
- $27=3^2+3^2+3^2$
Are these the only two such numbers to exist?
There has to be an easy proof for this, but it seems to elude me.
Thanks

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4=2^2

9=3^2

25=5^2

36=3^2 +3^2 + 3^2 +3^2

49=7^2

64=16*(2^2) or 4*4^2 or 3*4^2 + 4*2^2 or ....

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