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 April 8th, 2014, 05:13 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 Thanks from agentredlum
 April 8th, 2014, 09:58 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra $$13^3\ =\ 2197\ =\ 9^2\,+\,46^2\ =\ 26^2\ +\ 39^2$$ Thanks from MarkFL and William Labbett
 April 8th, 2014, 12:03 PM #3 Member   Joined: Apr 2014 From: norwich Posts: 84 Thanks: 9 Thanks for that. I am very impressed. Thanks
April 8th, 2014, 09:20 PM   #4
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Quote:
 Originally Posted by William Labbett does anyone know if any other prime other than 5, when cubed equals the sum of two distinct squares ?
I believe all primes = 1 mod 4 have this property. 17^3 = 47^2 + 52^2, for example.

 April 8th, 2014, 09:20 PM #5 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 234 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! Thanks from Hoempa, MarkFL, johnr and 1 others Last edited by agentredlum; April 8th, 2014 at 09:37 PM. Reason: Changed wording of 'ambitious' 1st sentence.
April 10th, 2014, 01:55 AM   #6
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Quote:
 Originally Posted by CRGreathouse I believe all primes = 1 mod 4 have this property. 17^3 = 47^2 + 52^2, for example.
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, 06:34 AM   #7
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 Originally Posted by mathsman1963 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.
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, 07:22 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 07:33 AM.
April 10th, 2014, 07:47 AM   #9
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Quote:
 Originally Posted by mathsman1963 There's an utterly bonny proof of this in Proofs from the Book, a collection of theorems inspired by Paul Erdos.
Great book.

Quote:
 Originally Posted by mathsman1963 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.
This problem is different from either of those, as I understand it. 0^2 is OK, but the squares can't be the same so 2 = 1^2 + 1^2 is disallowed.

 April 23rd, 2014, 12:06 PM #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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