April 8th, 2014, 04:13 AM  #1 
Member Joined: Apr 2014 From: norwich Posts: 84 Thanks: 9  cube of prime equal to sum of two distinct squares
Hi, does anyone know if any other prime other than 5, when cubed equals the sum of two distinct squares ? 5^3 = 125 11^2 + 2^2 = 125 also 10^2 + 5^2 = 125 
April 8th, 2014, 08:58 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra 
$$13^3\ =\ 2197\ =\ 9^2\,+\,46^2\ =\ 26^2\ +\ 39^2$$ 
April 8th, 2014, 11:03 AM  #3 
Member Joined: Apr 2014 From: norwich Posts: 84 Thanks: 9 
Thanks for that. I am very impressed. Thanks 
April 8th, 2014, 08:20 PM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  
April 8th, 2014, 08:20 PM  #5 
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 
You can generate many solutions to this problem if you write it this way , $$N^3 = a \cdot N^2 + b \cdot N^2 $$ $$N^3 = N^2 (a + b)$$ So , you want $$1) N = prime $$ $$2) \ a + b = N$$ $$3) \ a \ne b$$ 4) $a$ and $b$ to be both squares so that they can be 'absorbed' by N^2 , like a sponge absorbs water. This greatly limits the possibilities for any particular N. So , which primes can be written as the sum of two distinct squares? Make a list of squares and add them pairwise (two at a time) then check for primality , $$1^2 , 2^2 , 3^2 , 4^2 , 5^2 , 6^2, 7^2 , 8^2 , ... $$ N THAT WORK $N = 1^2 + 2^2 = 5 \ \ $ (your example) $N = 2^2 + 3^2 = 13 \ \ $ (Olinguito's example) $N = 1^2 + 4^2 = 17 \ \ $ (my example *and CRG's*) $N = 2^2 + 5^2 = 29 \ \ $ $N = 1^2 + 6^2 = 37 \ \ $ . . . Illustration of agent's 'absorption' method by worked out example Consider: $2^2 + 5^2 = 4 + 25 = 29$ $$29^3 = 4 \cdot 29^2 + 25 \cdot 29^2 $$ $$29^3 = 2^2 \cdot 29^2 + 5^2 \cdot 29^2 $$ $$29^3 = (2 \cdot 29)^2 + (5 \cdot 29)^2 $$ $4$ , $25$ have been 'absorbed' $$29^3 = 58^2 + 145^2$$ Have fun getting more working examples! Last edited by agentredlum; April 8th, 2014 at 08:37 PM. Reason: Changed wording of 'ambitious' 1st sentence. 
April 10th, 2014, 12:55 AM  #6 
Newbie Joined: Mar 2014 From: Manchester, United Kingdom Posts: 11 Thanks: 4 Math Focus: Number Theory  This is true: All primes of the form 4k+1 are expressible as the sum of two squares. Any number which is the product of numbers expressible as the sum of two squares is also expressible as the sum of two squares.

April 10th, 2014, 05:34 AM  #7 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Yes. You can take the product of any number of primes of the form 4n+1 and write them as the sum of two squares and hence (since the product is odd) a sum of two distinct squares. You can then multiply the result by any nonzero square and the result holds. The only question is how to handle numbers where 2 appears to an odd power. I *think* that these are OK exactly when there is some 4n+1 prime dividing the product (or maybe only with an even exponent?), but I have neither attempted proof nor studied it numerically.

April 10th, 2014, 06:22 AM  #8 
Newbie Joined: Mar 2014 From: Manchester, United Kingdom Posts: 11 Thanks: 4 Math Focus: Number Theory 
The complete theory for numbers which are the sum of two squares is that a number is expressible as the sum of two squares if and only if its prime divisors of the form 4k+3 have an even power in the number's canonical form. In this, 0^2 is permissible as one of the squares. If you want to restrict the squares to strictly positive ones then any powers of 2 in the canonical form must be odd. There's an utterly bonny proof of this in Proofs from the Book, a collection of theorems inspired by Paul Erdos. Proofs from THE BOOK  Wikipedia, the free encyclopedia Last edited by mathsman1963; April 10th, 2014 at 06:33 AM. 
April 10th, 2014, 06:47 AM  #9  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Quote:
 
April 23rd, 2014, 11:06 AM  #10 
Member Joined: Apr 2014 From: norwich Posts: 84 Thanks: 9 
Thanks very much for all the replies. I've not taken a look at Proof's from the book (yet). I can't see why a number which is the product of three primes, 1 of which is congruent to 1 modulo 4, the other 2 congruent to 3 moulo 4, can't be the sum of two squares, (which seems to be true from what's been said). I'm going to try to see if I can't find out myself. I'll check the book if I think that I can't solve it. 

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