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What Ramanujan had, every one should have

mathworker

Active member
May 31, 2013
118
hello we all are familiar with Ramanujan's number 1729 i thought of every one recognizing a special number well thought of mine ....65... (Wait).what so special about it (maybe not so special(Angry)):) it is smallest number to be expressed as sum of squares of two distinct pairs of numbers (i am not sure although!!) \(\displaystyle 1^2+8^2 = 4^2+7^2 \)
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: when Ramanujan have every one should have

Well, I don't have a certain number but I like perfect numbers which are the sum of its proper divisors such as (1+2+3=6) .
If I understand correctly 65 is no the least number that is the sum of two non-zero squares. It should be (1^2+2^2 = 5 ) .
 

mathworker

Active member
May 31, 2013
118
Re: when Ramanujan have every one should have

Well, I don't have a certain number but I like perfect numbers which are the sum of its proper divisors such as (1+2+3=6) .
If I understand correctly 65 is no the least number that is the sum of two non-zero squares. It should be (1^2+2^2 = 5 ) .
i am actually saying 65 is the least number which can be expressed as sum of squares of two numbers in two different ways like 1729=1^3+9^3=9^3+10^3...
65=1^2+8^2=4^2+7^2
 

agentmulder

Active member
Feb 9, 2012
33
Re: when Ramanujan have every one should have

hello we all are familiar with Ramanujan's number 1729 i thought of every one recognizing a special number well thought of mine ....65... (Wait).what so special about it (maybe not so special(Angry)):) it is smallest number to be expressed as sum of squares of two numbers (i am not sure although!!) 1^2+8^=4^2+7^2
Are we allowed to use the same number twice?

[tex]1^2 + 7^2 = 5^2 + 5^2 [/tex]

:)
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Can we find \(\displaystyle k^2+l^2 = m^2+n^2 \) are all distinct greater than 1 integers ?
 

mathworker

Active member
May 31, 2013
118

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Given that (5, 12, 13) and (12, 35, 37) are Pythagorean triples, we also have:

\(\displaystyle 13^2+35^2=5^2+37^2\)
 

TheBigBadBen

Active member
May 12, 2013
84
I found this paper, with some interesting relevant results

http://www.cs.toronto.edu/~mackay/sumsquares.pdf

as mathworker and agentmulder point out, the lowest number expressible as the sum of two squares in two different ways is
$$50 = 1^2 + 7^2 = 5^2 + 5^2$$
if the squares need not be distinct, or
$$65 = 1^2 + 8^2 = 4^2 + 7^2$$
otherwise.

What is interesting is that according to the paper, it is likely that there is some largest number n that is expressible as the sum of two squares in only one way.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
What is interesting is that according to the paper, it is likely that there is some largest number n that is expressible as the sum of two squares in only one way.
Not true. In fact, every prime number of the form $4n+1$ is uniquely expressible as a sum of two squares, and there are infinitely many such primes.

Here's a related problem. What is the smallest number that can be expressed as the sum of two squares in three different ways? My candidate is $$650 = 19^2+17^2 = 23^2+11^2 = 25^2+5^2,$$ but maybe there is a a smaller example?
 

agentmulder

Active member
Feb 9, 2012
33
Not true. In fact, every prime number of the form $4n+1$ is uniquely expressible as a sum of two squares, and there are infinitely many such primes.

Here's a related problem. What is the smallest number that can be expressed as the sum of two squares in three different ways? My candidate is $$650 = 19^2+17^2 = 23^2+11^2 = 25^2+5^2,$$ but maybe there is a a smaller example?
If we allow 1 again then

[tex] 325 = 1^2 + 18^2 [/tex]

[tex] 325 = 6^2 + 17^2 [/tex]

[tex] 325 = 10^2 + 15^2 [/tex]

But i can't be certain 325 is the smallest positive integer to be expressble in this way.

There is an interesting connection to COMPLEX NUMBERS for the generalisation of this problem. I'm sure Zaid will like that idea since he's so adept at working in the COMPLEX PLANE.

I'm sure other members can elaborate on this better than i can.

BTW,

[tex] 1105 = 4^2 + 33^2 = 9^2 + 32^2 = 12^2 + 31^2 = 23^2 + 24^2 [/tex]

None of them use 1^2

:)

P.S What TheBigBadBen said is not against what Opalg said. Primes of the form 4m + 1 are gauranteed to have 1 and only 1 representation as sum of 2 integer squares (<-- if i'm wrong about this i apologize), but I think TheBigBadBen is talking about non prime integer n as an upper limit. Which is indeed interesting... so , if i understand correctly, there exists non prime integer n = a^2 + b^2 with only one representation and for any non prime integer k > n if k has 1 representation then k must have at least 2 distinct representations.

I can't read the paper because my PS3 browser does not decode pdf. Long story...

:)
 
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agentmulder

Active member
Feb 9, 2012
33
Given that (5, 12, 13) and (12, 35, 37) are Pythagorean triples, we also have:

\(\displaystyle 13^2+35^2=5^2+37^2\)
This is interesting to me because you can draw the (5, 12, 13) right triangle and then construct the (12, 35, 37) right triangle upon one leg of
the (5, 12, 13) right triangle thereby getting a 'nice' visual of the problem. Can this be generalised?

Also interesting to me is 5, 13, 37 are all primes of the form 4m + 1, so each of 5, 13, 37 are expressible as sums of 2 distinct squares.

12 = 2^2*3 so neither prime factor is of the form 4m + 1, the prime factor 3 is of the form 4m + 3 but it does not occur to an even power so 12 is not expressible as the sum of 2 squares. (See further below for clarification of this point)

35 = 5*7 with the prime factor 7 of the form 4m + 3 but not to an even power so according to Fermat , Lagrange, etc. 35 is not expressible as the sum of 2 squares.

Quote from Wolfram Mathworld ...

'A positive integer can be represented as the sum of two squares iff each of its prime factors of the form 4k + 3 occurs as an even power, as first established by Euler in 1738. In Lagrange's four-square theorem, Lagrange proved that every positive integer can be written as the sum of at most four squares, although four may be reduced to three except for numbers of the form 4^n(8k + 7).'

From link below

Sum of Squares Function -- from Wolfram MathWorld

:)
 
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agentmulder

Active member
Feb 9, 2012
33
I used MarkFL's idea and my visual idea to find Pythagorean Triples

(20, 21. 29) , (21, 28 , 35)

From which we can get

[tex] 28^2 + 29^2 = 35^2 + 20^2 = 1625[/tex]

As a bonus we also get

[tex] 40^2 + 5^2 = 1625 [/tex]

Are there more bonuses for this particular example?

:)

See how number theory can be fun?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
29^2 + 28^2 = 1625
35^2 + 20^2 = 1625
37^2 + 16^2 = 1625
40^2 + 5^2 = 1625
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,707
... so , if i understand correctly, there exists non prime integer n = a^2 + b^2 with only one representation and for any non prime integer k > n if k has 1 representation then k must have at least 2 distinct representations.
No that's not true either. If you take a prime of the form $4n+1$ and multiply it by the square of a prime of the form $4m+3$, then that will have a unique expression as the sum of two squares. In particular, all numbers of the form $9(4n+1)$ (with $4n+1$ prime) will be of this form. Those numbers are not prime and there are infinitely many of them.
 

agentmulder

Active member
Feb 9, 2012
33
Here is my visual aid to MarkFL's idea. The line from the top vertex is perpendicular to the base making 2 right triangles. Denote the hypotenuse on the left as m^2 + n^2 as Euclid recommends, the bottom left leg as m^2 - n^2, the common leg as 2mn. Denote the hypotenuse on the right as k^2 + l^2, the bottom right leg as k^2 - l^2, the common leg as 2kl

2mn = 2kl

mn = kl

Now you're ready to plug in sme numbers and play. Use MarkFL's example (5, 12, 13). Put those numbers in the right triangle on the left with 12 as the common leg

2mn = 12 = 2kl

mn = 6 = kl

3*2 = 6 = 6*1

BEHOLD! There is your kl to give the other 2 sides.

k^2 - l^2 = 6^2 - 1^2 = 35

k^2 + l^2 = 6^2 + 1^2 = 37

So we generate MarkFL's example (5, 12, 13 ) , (12, 35, 37)

Notice corresponding side symmetry of 13^2 + 35^2 = 5^2 + 37^2.

For my example (20, 21, 29) we have to make a slight modification of this scheme, in particular, the common leg will be denoted as m^2 - n^2

Well, i hope you play around with this and see what you can come up with. Too bad i have limited graphing capability so i can't put the proper denotations on the sides myself. :)
 

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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
up to a certain number of digits it seems that the set of numbers that have maximum different representations of sums of squares always have numbers that are divisible by 5 .

up to 100000 the only number with 9 distinct representations is 93925

226^2 + 207^2= 93925
255^2 + 170^2= 93925
262^2 + 159^2= 93925
270^2 + 145^2= 93925
278^2 + 129^2= 93925
289^2 + 102^2= 93925
303^2 + 46^2= 93925
305^2 + 30^2= 93925
306^2 + 17^2= 93925

The maximum I reached is 204425 with 12 different representations

331^2 + 308^2= 204425
355^2 + 280^2= 204425
380^2 + 245^2= 204425
395^2 + 220^2= 204425
404^2 + 203^2= 204425
413^2 + 184^2= 204425
424^2 + 157^2= 204425
437^2 + 116^2= 204425
445^2 + 80^2= 204425
448^2 + 61^2= 204425
451^2 + 32^2= 204425
452^2 + 11^2= 204425
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
For big numbers not only they are divisble by 5 they are also divisble by 25 or any multiple of 5 .

655^2 + 610^2= 801125
703^2 + 554^2= 801125
710^2 + 545^2= 801125
722^2 + 529^2= 801125
766^2 + 463^2= 801125
769^2 + 458^2= 801125
785^2 + 430^2= 801125
815^2 + 370^2= 801125
830^2 + 335^2= 801125
862^2 + 241^2= 801125
865^2 + 230^2= 801125
874^2 + 193^2= 801125
881^2 + 158^2= 801125
886^2 + 127^2= 801125
890^2 + 95^2= 801125
895^2 + 10^2= 801125

It seems to always grow without bound .
 

TheBigBadBen

Active member
May 12, 2013
84