Numbers that are the sum of the squares of their prime factors?
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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1 Expert Answer
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 ....
Zaphod B.
The question asks for "prime factors with multiplicity". This means that 4=2*2 and 2^2+2^2=8 Also 49=7*7 and 7^2+7^2=98 Also 36=2*2*3*3 and 2^2+2^2+3^2+3^2=26 etc So, none of your examples is correct.
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11/12/21
Steven G.
tutor
Correct, none of my examples are correct as I did not give any examples. I showed the OP how to generates perfect squares and left it to them to finds results that satisfied their conjecture.
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11/12/21
Zaphod B.
I think that generating perfect squares is irrelevant to this question... You just have to factor the number and sum the squares of the prime factors. What the OP is asking is much deeper. He asks if the Markoff–Hurwitz equation for a=1 has solutions on primes. This is a well known problem which was recently answered by Giorgos Kalogeropoulos. I posted the details about the answer as a comment at the main question.
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11/15/21
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Zaphod B.
Actually, those 2 are not the only ones! Giorgos Kalogeropoulos has found 3 more numbers, each having more than 100 digits. You can find these numbers if you follow the links in the comments of OEIS A339062 & A338093 or here (where the original answer was posted): https://www.primepuzzles.net/puzzles/puzz_1019.htm So, such numbers exist! It is an open question if there are infinitely many of them...11/12/21