Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Equation: $X^2+Y^2=Z^2+1$
How to find all positive integer solutions to this equation.


Can you find $n\in \mathbb{N}$ such that $X^2+Y^2=Z^2+n$ has no positive integers as solutions ?

share|improve this question
3  
Please be polite and do not post in the imperative. What have you tried? –  user38268 Aug 25 '11 at 6:37

2 Answers 2

up vote 5 down vote accepted

There is no such $n$. Rewrite as $y^2-n=z^2-x^2$. Choose $y$ so left side is positive and not twice an odd number (always possible: just take $y$ of parity opposite to $n$), every such number can be written as a difference of 2 squares. Even easier: let $x=(n^2+n)/2$, $y=n+1$, $z=(n^2+n+2)/2$.

EDIT: Concerning the first question, some of the discussion at this MathOverflow question might be of interest.

share|improve this answer

Gerry Myerson has dealt with the second question. I will say a few words about the first. The ring of Gaussian Integers, that is $\mathbb{Z}[i],$ is the natural place to address this question. Many algebra texts use this ring to prove that a prime $p \equiv 1$ (mod 4) is a sum of two (rational integer) squares. The ring of Gaussian integers, is a Euclidean domain, hence a unique factorization domain. Its arithmetic is complicated somewhat by the existence of the four units $\{1,-1,i,-i\},$ and uniqueness of prime factorizations is only up to reordering of terms and multiplication of terms by a unit. It has three kinds of prime: rational primes $p \equiv 3$ (mod 4) remain prime in $\mathbb{Z}[i],$ primes $p \equiv 1$ (mod 4) are expressible as a product of two different primes in $\mathbb{Z}[i],$ so when $p \equiv 1$ (mod 4), we can always write $p = (a+bi)(a-bi)$ for coprime (rational) integers $a$ and $b$ (note that also $p = (b+ai)(b-ai)$). In this case, we have ${\rm gcd}(a+bi,a-bi) = 1$ in $\mathbb{Z}[i].$ The prime $2$ is very special in $\mathbb{Z}[i].$ We have $2 = (1+i)(1-i) = -i(1+i)^2,$ so $2$ is a unit times the square of a prime element of $\mathbb{Z}[i].$

To solve $X^2 + Y^2 = Z^2 + 1$ in rational integers, we note the following: (*) For any even integer $Z$, the integer $Z^2 + 1$ is a product of primes which are each congruent to $1$ (mod $4$). For note that $Z^2 + 1$ is odd, so that neither $Z+i$ nor $Z -i$ can have a factor $1+i$ in $\mathbb{Z}[i]$ (if one did, taking complex conjugates shows that the other one does, since $1+i$ divides $1-i$). If $p$ were a prime congruent to $3$ (mod 4), which divided $Z^2+1$, then in $\mathbb{Z}[i],$ we have $p$ divides $(Z+i)(Z-i),$ so since $p$ remains prime in the Euclidean domain $\mathbb{Z}[i],$ we have $p$ divides $Z+i$ or $p$ divides $Z-i.$ But, taking complex conjugates, if it divides one, then it divides the other. Then $p$ divides $Z+i -(Z-i),$ so that $p$ divides $2i$, a contradiction (this is one way that $2$ behaves differently).

(**) For any odd integer $Z$, the integer $Z^2+1$ is divisible by $2$, but not $4,$ and its remaining (rational) primes factors are all congruent to $1$ (mod 4). The statement about odd prime factors follows as above, and we note that $Z^2+1$ is congruent to $2$ (mod 4) when $Z$ is odd, so $Z^2+1$ is divisible by $2$, but not by $4,$ in that case.

Returning to the original question, there are two cases to consider: if $Z$ is even, we know we can factorize $Z^2 + 1$ in the form $p_1^{m_1} \ldots p_n^{m_n}$ where the $p_j$ are distinct (rational) primes, each congruent to $1$ (mod 4). For each $j,$ we may write $p_j = \pi_j \overline{\pi_{j}},$ where each $\pi_j = a_j + ib_j$ for rational integer $a_j$ and $b_j$, and each $\pi_j$ is a prime element of $\mathbb{Z}[i].$

If $Z$ is odd, we can factorize $Z^2 + 1$ in the form $2.p_1^{m_1} \ldots p_n^{m_n}$ where the $p_j$ are distinct (rational) primes, each congruent to $1$ (mod 4). For each $j,$ we may write $p_j = \pi_j \overline{\pi_{j}},$ where each $\pi_j = a_j + ib_j$ for rational integer $a_j$ and $b_j$, and each $\pi_j$ is a prime element of $\mathbb{Z}[i].$ We can also write $2 = \pi_0 \overline{\pi}_0,$ where $\pi_0 = (1+i)$.

Hence we have (in $\mathbb{Z}[i]$), a prime factorization of $(Z+i),$ which is as unique as it can be, which , (possibly after switching $\pi_j$ and $\overline{\pi}_j$ and relabelling), has the form $Z+i = \pi_0 ^{\delta} \prod_{j=1}^{n} \pi_{j}^{m_j},$ where $\delta = 0$ if $Z$ is even, and $\delta = 1$ if $Z$ is odd.

Then a way to solve $X^2 + Y^2$ in rational integers is to write $X^2 + Y^2 = (X+iY)(X-iY)$ in $\mathbb{Z}[i].$ If we factorize $X + iY$ as a proudct of primes in $\mathbb{Z}[i],$ we get a prime factorization of $X-iY$ by complex conjugation, and then juxtaposing them gives a prime factorization of $Z^2 + 1,$ so we can exploit the uniqueness.

Up to multiplication by a unit, we must have $X + iY = \pi_{0}^{\delta} \prod_{j = 1}^{n} \pi_{j}^{r_j}\overline{\pi}_{j}^{s_j}$, where for each $j,$ we have $r_j \leq m_j$ and $s_j \leq m_j.$ Taking complex conjugates gives $X-iY$ in a similar form. Now we notice that for each $j$, we must have $r_j + s_j =m_j,$ to get the right prime factorization for $Z+i.$

Then we use $2X = (X+iY) +(X-iY)$ and $2Y= -i(X+iY) +i(X-iY)$ to calculate $X$ and $Y$.

In summary, all solutions can be found as follows: choose any positive integer $Z$, and factorize $Z+i$ in $\mathbb{Z}[i]$. This factorization will (up to multiplication by a unit) have the form $Z+i = \pi_0 \prod_{j=1}^{n} \pi_j ^{m_j}$, where $\pi_0 = i+i$ and $\delta = 0,1$ if $Z$ is even, odd respectively, and where there are $n$ distinct (rational) primes $p_j,$ each congruent to $1$ (mod 4), such that $p_j = |\pi_j|^2$ for each $j,$ each $\pi_j$ being a Gaussian integer. We may then choose $X+iY$ to be any expression of the form $\pi_0^{\delta} \prod_{j=1}^{n} \pi_{j}^{r_j} \overline{\pi_j}^{m_j - r_j}$, where $0 \leq r_j \leq m_j$ for each $j.$ Since you insist that $X$ and $Y$ be positive, it may be necessary to replace $X$ by $-X$, etc.

share|improve this answer
    
I imagined OP to be asking for parametric formulas giving all solutions, like the formulas $\lbrace x,y\rbrace=\lbrace2kmn,k(m^2-n^2)\rbrace$, $z=k(m^2+n^2)$ for $x^2+y^2=z^2$. I don't know if there are any. –  Gerry Myerson Aug 25 '11 at 12:55
    
But your link which you edited in to your answer in seems to suggest (strongly) that there is a parametrization. –  Geoff Robinson Aug 26 '11 at 23:47
    
There is a parametrization, but it's a bit of a cheat, as the parameters are related by $rt+su=2$, and how do you parametrize that? –  Gerry Myerson Aug 27 '11 at 1:15
    
That parametrization and what I am doing seem closely related, come to think of if. Factoring $z+i$ as $(r+si)(t+ui)$ in $\mathbb{Z}[i]$ amounts to writing $z = rt-us$ where $ru+st = 1.$ –  Geoff Robinson Aug 27 '11 at 13:43
    
Yes, quite.${}$ –  Gerry Myerson Aug 27 '11 at 22:42

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.