sum of squares of two rational numbers

[logistic_guy]

here is the question

Determine the number of ways to write each of the following rational integers as the sum of squares of two rational integers.

(a) \(\displaystyle 5\)
(b) \(\displaystyle 20\)
(c) \(\displaystyle 120\)
(d) \(\displaystyle 1000\)


my attemb
by trial and error i find \(\displaystyle 2^2 + 1^2 = 4 + 1 = 5\)
\(\displaystyle 2^2 + 4^2 = 4 + 16 = 20\)
i can't find sum for \(\displaystyle 120\)
i can't find sum for \(\displaystyle 1000\)
is my solution correct?

 

[BeansNRice]

well obviously the two you did are correct.

I don't see any definitive way to solve these as there is a single equation and two unknowns.

Probably your best bet is just run through the obvious squares, subtract that from the given number and see which leaves another square.

120 doesn't appear to have a solution

1000 has two solutions! Keep trying!

As an aside, your terminology "rational integer" must be in error. All integers are rational, but not all rationals are integers. Based on the nature of the problem I'd guess that the term you are looking for is just "integer".

 

[Dr.Peterson]

As an aside, your terminology "rational integer" must be in error. All integers are rational, but not all rationals are integers. Based on the nature of the problem I'd guess that the term you are looking for is just "integer".

No, the term "rational integer" is used, though not usually in this context where the meaning is obvious:

Rational Integer -- from Wolfram MathWorld

A synonym for integer. The word "rational" is sometimes used for emphasis to distinguish it from other types of "integers" such as cyclotomic integers, Eisenstein integers, Gaussian integers, and Hamiltonian integers.

mathworld.wolfram.com

But in dropping the word "integer" from the title, the meaning was changed significantly.

Determine the number of ways to write each of the following rational integers as the sum of squares of two rational integers.

(a) \(\displaystyle 5\)
(b) \(\displaystyle 20\)
(c) \(\displaystyle 120\)
(d) \(\displaystyle 1000\)

I don't know enough about Diophantine equations (or number theory in general) to see a quicker way than trial and error; and Google's AI gave only a trial and error answer; but this theorem appears to be relevant:

Sum of two squares theorem - Wikipedia

en.wikipedia.org

 

[fresh_42]

From [imath] N=r^2+s^2 [/imath] with [imath] r,s\in \mathbb{Q} [/imath] we get an integer equation [imath] c^2\cdot N=a^2+b^2. [/imath] If [imath] p^{2k+s} [/imath] divides [imath] N [/imath] then we can put [imath] p^{k} [/imath] into [imath] c [/imath] and still have an equation of the form [imath] c^2N=a^2+b^2 [/imath] but this time with only [imath] p^s [/imath] that divides [imath] N. [/imath] If we do this with all prime factors of [imath] N, [/imath] we get an equation [imath] c^2N=a^2+b^2 [/imath] where [imath] N [/imath] is square-free, i.e. [imath] N=p_1\cdots p_n [/imath] with pairwise distinct primes [imath] p_1,\ldots,p_n. [/imath]

In your example of [imath] N=1000, [/imath] we would turn [imath] c^2\cdot 1000=a^2+b^2 [/imath] into [imath] (10\cdot c)^2\cdot 10=a^2+b^2.[/imath]

The general question is therefore: Is there a solution for the integer equation [imath] c^2N=a^2+b^2 [/imath] such that [imath] N= p_1\cdots p_n[/imath] with pairwise distinct primes [imath] p_1,\ldots,p_n. [/imath] This means we must investigate [imath] c^2\cdot p_1\cdots p_n=a^2+b^2. [/imath]

There are two helpful results in this context:

a) A product of sums of two squares is again a sum of two squares: [imath] (a_1^2+b_1^2)\cdots (a_k^2+b_k^2)=a^2+b^2. [/imath]

b) An odd prime [imath] p [/imath] can be written as [imath] p= x^2+y^2[/imath] with integers [imath] x,y [/imath] if and only if [imath] 4 [/imath] divides [imath] p-1. [/imath]

I'm a bit in a hurry now, and cannot see how this helps.

 

[logistic_guy]

thank Beans

120 doesn't appear to have a solution

how do you decide \(\displaystyle 120\) have no solutution? do you check all the numbers?

I don't know enough about Diophantine equations (or number theory in general) to see a quicker way than trial and error; and Google's AI gave only a trial and error answer; but this theorem appears to be relevant:

Sum of two squares theorem - Wikipedia

en.wikipedia.org

thank Dr.

i'll read it

There are two helpful results in this context:

a) A product of sums of two squares is again a sum of two squares: [imath] (a_1^2+b_1^2)\cdots (a_k^2+b_k^2)=a^2+b^2. [/imath]

b) An odd prime [imath] p [/imath] can be written as [imath] p= x^2+y^2[/imath] with integers [imath] x,y [/imath] if and only if [imath] 4 [/imath] divides [imath] p-1. [/imath]

I'm a bit in a hurry now, and cannot see how this helps.

thank fresh_42

i don't understand a)
b) appear i can get solution with factoring the number into primes numbers

\(\displaystyle 5\) itself prime, it have power \(\displaystyle 1\) as \(\displaystyle 5^1\)
is there anything especial about the odd power?
i can also see \(\displaystyle \frac{4}{5 - 1} = \frac{4}{4} = 1\)
are you trying to say if this true, then \(\displaystyle 5\) can be written with two square like i do \(\displaystyle 2^2 + 1^2 = 5\)

 

[logistic_guy]

but this theorem appears to be relevant:

Sum of two squares theorem - Wikipedia

en.wikipedia.org

thank Dr. very much

i like the Jacobi's two-square theorem. i think it's useful but don't fully understand it

i'll work with the prime \(\displaystyle 5\)

\(\displaystyle n = 2^f p_1^{r_1} p_2^{r_2}\cdots q_1^{s_1}q_2^{s_2} = 2^05^1 = 5\)

\(\displaystyle 2^f = 2^0 = 1\)
\(\displaystyle p_1^{r_1} = 5^1 = 5\)
what is \(\displaystyle p_2^{r_2}\)?
what is \(\displaystyle q_1^{s_1}\)?
what is \(\displaystyle q_2^{s_2}\)?

\(\displaystyle p_1 \equiv 5 \equiv 1 \ mod \ 4\)
\(\displaystyle q_1 \equiv \ ? \equiv 3 \ mod \ 4\)

what is \(\displaystyle r_2(n) = \ ?\) in this case?

 

[fresh_42]

i don't understand a)

It is not obvious, it needs a proof. I can link to one if you like but you would need to download a pdf. Or I can write it here, but that will take some time and I'm not sure whether it is worth the trouble with the MathJax here.

b) appear i can get solution with factoring the number into primes numbers

\(\displaystyle 5\) itself prime, it have power \(\displaystyle 1\) as \(\displaystyle 5^1\)
is there anything especial about the odd power?

No, I only tried to simplify the problem as far as possible, and
[math] c^2p_1\cdots p_n=a^2+b^2 [/math]is as far as I got by a quick examination.

My next step will be trying to rule out [imath] p_1=2 [/imath] so I may assume that all prime factors are odd.

Fermat has proven my second statement that [imath] p=a^2+b^2 \Longleftrightarrow p\equiv 1\pmod{4} [/imath] for odd primes.

This rules out the case that all [imath] p_k\equiv 1\pmod{4} [/imath] and we may assume that [imath] p_1\equiv 3\pmod{4}. [/imath] I hope to rule out this case, too.

That would be my plan.

 

[logistic_guy]

It is not obvious, it needs a proof. I can link to one if you like but you would need to download a pdf. Or I can write it here, but that will take some time and I'm not sure whether it is worth the trouble with the MathJax here.

you don't have to write and waste your time.

i'm just want if there's simple formula or theorem like the Jacobi. it can tell if a number have sums. i just can't understand how to apply it to the numbers \(\displaystyle 5, 20, 120, 1000\)

it look simple but the notation is new. i think if i can apply it to \(\displaystyle 5\), i'll be able to apply it to all other number

 

[fresh_42]

Well, you can make them square-free, so we have [imath] 5\, , \,2^2\cdot 5\, , \, 4\cdot 30=2^2\cdot 2\cdot 3\cdot5\, , \,10^2\cdot 10=10^2\cdot 2\cdot 5 .[/imath] Since we have rational numbers here, we can divide the square and are left with the situation [imath] 5\, , \,30\, , \,10. [/imath] Since [imath] 5=1^2+2^2 [/imath] we are left with the demand to write [imath] 6 [/imath] and [imath] 2 [/imath] as square of primes.

[imath] 2=1^2+1^2 [/imath] so [imath] 20=2^2(1^2+2^2) [/imath] and [imath] 1000= 10^2(1^2+1^2)(1^2+2^2)[/imath] which is again of the form [imath] a^2+b^2 [/imath] by the statement we spoke about. Hence, [imath] 5,20,1000 [/imath] can be written as such.

Remains to prove, that [imath] 120=2^2(1^2+2^2)(1^2+1^2)\cdot 3 [/imath] can not. Again, we may divide the equation by [imath] 2^2(1^2+2^2)(1^2+1^2) [/imath] because we have rational numbers. That leaves us with the question of whether [imath] 3=x^2+y^2 [/imath] is solvable over the rational numbers or not.

But this can be done by looking at the remainders by a division by [imath] 3. [/imath] Say
[math] 3=\dfrac{p^2}{q^2}+\dfrac{r^2}{s^2}\Longrightarrow 3\,q^2s^2=p^2+r^2\Longrightarrow 3\,|\,p^2+r^2[/math]Square integers are either divisible by [imath] 3 [/imath] or have a remainder [imath] 1. [/imath] Since [imath] 1+1=2\not\equiv 0\pmod{3} [/imath] and [imath] 1+0=1\not\equiv 0\pmod{3} ,[/imath] we have [imath] 3\,|\,p [/imath] and [imath] 3\,|\,r. [/imath] This leaves us with an even number of threes on the right and an odd number of threes on the left which is impossible. Hence [imath] 3 [/imath] cannot be written as a sum of squares (which also follows from Fermat's statement) and therefore [imath] 120 [/imath] cannot be written as a sum of square rational numbers either.

 

[BeansNRice]

this place kills me... question appears in Beginner Algebra forum and you all are off into fairly deep number theory.

 

[fresh_42]

this place kills me... question appears in Beginner Algebra forum and you all are off into fairly deep number theory.

Yes, but the condition rational numbers rules out induction, and there are really many of them of all sizes. Proving that something is impossible is generally hard.

 

[Dr.Peterson]

this place kills me... question appears in Beginner Algebra forum and you all are off into fairly deep number theory.

This particular OP posts everything at this level or lower, and is all over the place in the level of his questions. (Look at his history of postings.) I have no idea what he is doing here, but it's clear that he either doesn't know the meaning of things like "Beginning Algebra", or just doesn't care; those of us who are familiar with his questions know to ignore what forum they are in (and, often, to ignore him entirely). I don't think he will benefit from the deep number theory, based on his demonstrated lack of understanding, but who knows? Let someone try.

And who do you mean by "all"? I'm not wasting my time on it.

Yes, but the condition rational numbers rules out induction, and there are really many of them of all sizes. Proving that something is impossible is generally hard.

As I pointed out, the actual question is about integers, not rational numbers. It's the title that is wrong.

But in any case, you are likely going beyond what the OP is prepared to follow. That's the real issue.

 

[fresh_42]

But in any case, you are likely going beyond what the OP is prepared to follow. That's the real issue.

Yes, I am no friend of knowledge reserved for any kind of elites. If answers cannot be understood, then people can post questions about it. However, in this case, it is all about primes and divisibility, not really hard to understand. That [imath] (a_1^2+b_1^2)\cdot \ldots \cdot (a_n^2+b_n^2) [/imath] can be written as [imath] a^2+b^2 [/imath] again is a useful tool and worth keeping in mind. It follows easily if we may use determinants, but an induction should do. The other statement I quoted was from Fermat. If it could be found in the 17th century, then it cannot be that hard. But I admit that Fermat's theorem is a bit tricky.

I thought this was a platform where people can learn something.

 

[logistic_guy]

Well, you can make them square-free, so we have [imath] 5\, , \,2^2\cdot 5\, , \, 4\cdot 30=2^2\cdot 2\cdot 3\cdot5\, , \,10^2\cdot 10=10^2\cdot 2\cdot 5 .[/imath] Since we have rational numbers here, we can divide the square and are left with the situation [imath] 5\, , \,30\, , \,10. [/imath] Since [imath] 5=1^2+2^2 [/imath] we are left with the demand to write [imath] 6 [/imath] and [imath] 2 [/imath] as square of primes.

[imath] 2=1^2+1^2 [/imath] so [imath] 20=2^2(1^2+2^2) [/imath] and [imath] 1000= 10^2(1^2+1^2)(1^2+2^2)[/imath] which is again of the form [imath] a^2+b^2 [/imath] by the statement we spoke about. Hence, [imath] 5,20,1000 [/imath] can be written as such.

Remains to prove, that [imath] 120=2^2(1^2+2^2)(1^2+1^2)\cdot 3 [/imath] can not. Again, we may divide the equation by [imath] 2^2(1^2+2^2)(1^2+1^2) [/imath] because we have rational numbers. That leaves us with the question of whether [imath] 3=x^2+y^2 [/imath] is solvable over the rational numbers or not.

But this can be done by looking at the remainders by a division by [imath] 3. [/imath] Say
[math] 3=\dfrac{p^2}{q^2}+\dfrac{r^2}{s^2}\Longrightarrow 3\,q^2s^2=p^2+r^2\Longrightarrow 3\,|\,p^2+r^2[/math]Square integers are either divisible by [imath] 3 [/imath] or have a remainder [imath] 1. [/imath] Since [imath] 1+1=2\not\equiv 0\pmod{3} [/imath] and [imath] 1+0=1\not\equiv 0\pmod{3} ,[/imath] we have [imath] 3\,|\,p [/imath] and [imath] 3\,|\,r. [/imath] This leaves us with an even number of threes on the right and an odd number of threes on the left which is impossible. Hence [imath] 3 [/imath] cannot be written as a sum of squares (which also follows from Fermat's statement) and therefore [imath] 120 [/imath] cannot be written as a sum of square rational numbers either.

what you write here is very nice but i want a rule or theorem to follow

i'll rewrite the Jacobi theorem and i'll enhance it

Jacobi's two-square theorem
Two-square theorem --- Denote the number of divisors of \(\displaystyle n\) as \(\displaystyle d(n)\), and write \(\displaystyle d_a(n)\) for the number of those divisors with \(\displaystyle d \equiv a \ mod \ 4\). Let \(\displaystyle n = 2^f p_1^{r_1} p_2^{r_2}\cdots q_1^{s_1}q_2^{s_2}\cdots\) where \(\displaystyle p_i \equiv 1 \ mod \ 4\), \(\displaystyle q_i \equiv 3 \ mod \ 4\).

Let \(\displaystyle r_2(n)\) be the number of ways \(\displaystyle n\) can be represented as the sum of two squares.
Then \(\displaystyle r_2(n) = 0\) if any of the exponents \(\displaystyle s_j\) are odd. If all \(\displaystyle s_j\) are even then
\(\displaystyle r_2(n) = 4d(p_1^{r_1}p_2^{r_2}\cdots) = 4(d_1(n) - d_3(n)) = (r_1 + 1) \cdots (r_k + 1)\)

i think i understand the theorem now. i'll use it to solve the numbers \(\displaystyle 5,20,120,1000\). if i make mistake stop me.

case 1
i ignore the even prime i think
1- this theorem say if the odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), check the power, if any is odd, stop, there is no sum
2- this theorem say if the odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), check the power, if all is even, there is sum
if second condition satisfy
***
take power of odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\) and \(\displaystyle r_2(n) = 4(r_1 + 1) \cdots (r_k + 1)\)

case 2
what happen if there's no odd primes at all?

\(\displaystyle 5 = 2^05^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(5) = 4(1 + 1) = 4(2) = 8\)
this mean there's eight way to write \(\displaystyle 5\) with square sum. the answer don't make sense and i can't poof it

\(\displaystyle 20 = 2^25^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(20) = 4(1 + 1) = 4(2) = 8\)

\(\displaystyle 120 = 2^33^15^1\)
there's odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) with odd power
so there's no sum

\(\displaystyle 1000 = 2^35^3\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(1000) = 4(3 + 1) = 4(4) = 16\)

i'm not sure what happen if there's two odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)

\(\displaystyle 47385000 = 2^35^413^13^6\)
\(\displaystyle 2^3:\) i ignore even primes
\(\displaystyle 3^6: \) all odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) have even power
\(\displaystyle 5\) and \(\displaystyle 13\) is odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
so \(\displaystyle r_2(47385000) = 4(4 + 1)(1 + 1) = 4(5)(2) = 4(10) = 40\)
this mean there's forty way to write \(\displaystyle 47385000\) with square sum

is my analize correct?

 

[fresh_42]

Let me see if I can follow you.

\(\displaystyle 5 = 2^05^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(5) = 4(1 + 1) = 4(2) = 8\)
this mean there's eight way to write \(\displaystyle 5\) with square sum. the answer don't make sense and i can't poof it

I think this only makes sense if we count different signs and order. We have
[math] 5=(-1)^2+(-2)^2=(-1)^2+2^2=1^2+(-2)^2=1^2+2^2=(-2)^2+(-1)^2=(-2)^2+1^2=2^2+(-1)^2=2^2+1^2 [/math]This is the only way I see to get eight. If we assume [imath] 0<a\leq b [/imath] then [imath]0< a^2+b^2\leq 2b^2. [/imath] This means that [imath]b^2< 5=a^2+b^2\leq 2b^2 [/imath] and [imath] b\ge 2. [/imath] Since [imath] 3^2>5 [/imath] we must have [imath] b=2 [/imath] and [imath] a=1. [/imath] So disregarding order and signs leaves us with only one solution.

\(\displaystyle 20 = 2^25^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(20) = 4(1 + 1) = 4(2) = 8\)

I guess, again by allowing signs and distinction between [imath] 2^2+4^2 [/imath] and

\(\displaystyle 120 = 2^33^15^1\)
there's odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) with odd power
so there's no sum

\(\displaystyle 1000 = 2^35^3\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(1000) = 4(3 + 1) = 4(4) = 16\)

i'm not sure what happen if there's two odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)

\(\displaystyle 47385000 = 2^35^413^13^6\)
\(\displaystyle 2^3:\) i ignore even primes
\(\displaystyle 3^6: \) all odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) have even power
\(\displaystyle 5\) and \(\displaystyle 13\) is odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
so \(\displaystyle r_2(47385000) = 4(4 + 1)(1 + 1) = 4(5)(2) = 4(10) = 40\)
this mean there's forty way to write \(\displaystyle 47385000\) with square sum

is my analize correct?

 

[fresh_42]

Sorry, my computer crashed before I finished the previous post and my edit slot timed out. So here we go again:

Let me see if I can follow you.

\(\displaystyle 5 = 2^05^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(5) = 4(1 + 1) = 4(2) = 8\)
this mean there's eight way to write \(\displaystyle 5\) with square sum. the answer don't make sense and i can't poof it

I think this only makes sense if we count different signs and order. We have
[math] 5=(-1)^2+(-2)^2=(-1)^2+2^2=1^2+(-2)^2=1^2+2^2=(-2)^2+(-1)^2=(-2)^2+1^2=2^2+(-1)^2=2^2+1^2 [/math]This is the only way I see to get eight. If we assume [imath] 0<a\leq b [/imath] then [imath]0< a^2+b^2\leq 2b^2. [/imath] This means that [imath]b^2< 5=a^2+b^2\leq 2b^2 [/imath] and [imath] b\ge 2. [/imath] Since [imath] 3^2>5 [/imath] we must have [imath] b=2 [/imath] and [imath] a=1. [/imath] So disregarding order and signs leaves us with only one solution.

\(\displaystyle 20 = 2^25^1\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(20) = 4(1 + 1) = 4(2) = 8\)

I guess, again by allowing signs and distinction between [imath] 2^2+4^2 [/imath] and [imath] 4^2+2^2. [/imath]

\(\displaystyle 120 = 2^33^15^1\)
there's odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) with odd power
so there's no sum

The reason is that [imath] d_1(n)-d_3(n)=1-1=0 [/imath] since [imath] 120=2^3\cdot 3^1\cdot 5^1. [/imath]

\(\displaystyle 1000 = 2^35^3\)
there's no odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\), i'm guess in this case we just continue from the three stars***
so \(\displaystyle r_2(1000) = 4(3 + 1) = 4(4) = 16\)

i'm not sure what happen if there's two odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)

We have [imath] 1000=100\cdot 10=10^2\cdot 2\cdot 5=10^2\cdot(1^2+1^2)\cdot (1^2+2^2) [/imath] and by the product formula, it can be written as a sum [imath] a^2+b^2. [/imath] Counting all possibilities with signs and order is troublesome. We have [imath] 1000=2^3\cdot 5^3 [/imath] so the formula is [imath] r_2(n)=4(d_1(n)-d_3(n))=4d_1(n). [/imath] I see four divisors [imath] 1,5,25,125 [/imath] as divisors modulo [imath] d\equiv 1\pmod{4} [/imath] and none [imath] d\equiv 3\pmod{4}. [/imath] Looks ok.

\(\displaystyle 47385000 = 2^35^413^13^6\)
\(\displaystyle 2^3:\) i ignore even primes
\(\displaystyle 3^6: \) all odd primes is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\) have even power
\(\displaystyle 5\) and \(\displaystyle 13\) is odd primes is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
so \(\displaystyle r_2(47385000) = 4(4 + 1)(1 + 1) = 4(5)(2) = 4(10) = 40\)
this mean there's forty way to write \(\displaystyle 47385000\) with square sum

is my analize correct?

We have [imath] 47385000=2^3\cdot 3^6\cdot 5^4\cdot 13 [/imath]

That's a whole lot of divisors to count to determine [imath] d_1(47385000) [/imath] and [imath] d_3(47385000). [/imath] I tried, but I gave up.

 

[logistic_guy]

Sorry, my computer crashed before I finished the previous post and my edit slot timed out. So here we go again:

Let me see if I can follow you.

thank fresh_42 veru much

i see now how the \(\displaystyle 8\) way turn into \(\displaystyle 1\) solution. the same thing apply to other numbers

i see the number \(\displaystyle 120\) has two divisors congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
\(\displaystyle 1,5\)
and one divisor congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\)
\(\displaystyle 3\)

i'm wrong?

 

[logistic_guy]

if i say so, \(\displaystyle d_1(n) - d_3(n) = 2 - 1 = 1\) which is wrong

 

[fresh_42]

You have to count all divisors, not just the primes. But the formula says that it only applies if all [imath] s_j [/imath] are even which is not the case here. We have an odd [imath] s_1=1 [/imath] so [imath] r_2(120)=0. [/imath] My argument was wrong, I made a mistake.

 

[logistic_guy]

You have to count all divisors, not just the primes.

\(\displaystyle 1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120\)

so

\(\displaystyle 1,5\) is congruent to \(\displaystyle 1\) modulo \(\displaystyle 4\)
\(\displaystyle 3,15\) is congruent to \(\displaystyle 3\) modulo \(\displaystyle 4\)
so
\(\displaystyle d_1(n) - d_3(n) = 2 - 2 = 0\)

is it correct?