square that can be written as mean of as many pairs of squares as possible

If $b^2=\displaystyle\frac{a_1^2+c_1^2}2$ and $b^2=\displaystyle\frac{a_2^2+c_2^2}2$... and $b^2=\displaystyle\frac{a_n^2+c_n^2}2$, what is the largest possible value of $n$, or can $n$ be arbitrarily large?

e.g for $n=7$, $$325^2=\frac{49^2+457^2}2=\frac{65^2+455^2}2=\frac{115^2+445^2}2=\frac{175^2+425^2}2$$ $$=\frac{221^2+403^2}2=\frac{235^2+395^2}2=\frac{287^2+359^2}2,$$ and $$425^2=\frac{7^2+601^2}2=\frac{85^2+595^2}2=\frac{175^2+575^2}2=\frac{205^2+565^2}2$$ $$=\frac{289^2+527^2}2=\frac{329^2+503^2}2=\frac{355^2+485^2}2.$$

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Taking norms of both sides of suitable factorizations in the ring of Gaussian integers $\mathbf{Z}[i]$ would explain these. $N(a+bi)=a^2+b^2$, $N(1+i)=2$, $N((a+bi)(c+di))=N(a+bi)N(c+di)$. Furthermore $p=N(a+bi)$ is solvable for any prime $p\equiv 1\pmod4.$ – Jyrki Lahtonen Dec 27 '11 at 8:18
Have you tried working over the complex numbers? Factoring $a^2 + b^2 = (a+bi)(a-bi)$ might be interesting, since you're actually looking for an arbitrary large possibility of factorizations into two complex conjugates of a number of the form $2n^2$. I believe there are many possibilities but I am not in shape, number-theoretically speaking, to work it out right now. – Patrick Da Silva Dec 27 '11 at 8:19

You are asking for an integer ($2b^2)$ that can be written as a sum of two squares in as many ways as possible, one of which is a sum of two equal squares. By the result that I cited in this answer, there is no bound to this number of ways. Concretely, for instance $2\times25^n$ can be written as a sum of two squares in $4(2n+1)$ ways; these include $4$ ways of the form $(\pm5^n)^2+(\pm5^n)^2$ that you don't want to count and the remaining $8n$ possibilities come in symmetry classes of $8$ (by signs and order of terms), for a total of $n$ classes.

In terms of your question, the square $(5^n)^2$ can be written as the average of two squares in $n$ non-equivalent ways. To find these expressions, take the Gaussian integers $1+\mathbf i$ and $2n$ copies of $2+\mathbf i$, conjugate $i$ of the latter for $0\leq i<n$ and multiply everything together; the resulting $n$ Gaussian integers are all of norm-squared equal to $2\times25^n$, and their real and imaginary parts provide $n$ non-equivalent pairs of numbers, the averages of whose squares is $25^n=(5^n)^2$.

This is not the most economic way to get lots of expressions; it would be better to combine distinct prime numbers congruent to $1$ modulo $4$, rather than to take powers of one of them namely $5$. This explains your examples $325=5^2\times13$ and $425=5^2\times17$.

Added: The general formula for the number of solutions for writing $N^2$ as the average of a set of two squares of distinct positive numbers, in terms of the prime factorization of $N$, is as follows: only the primes congruent to $1$ modulo $4$ contribute; multiply together for every nonzero multiplicity $m$ of such a prime the numbers $2m+1$, subtract $1$ from the the (odd) product so obtained, and divide by $2$ (which accounts for the ignored order). So for $N=325=5^2\times13$ and $N=425=5^2\times17$ one gets $\frac{5\times3-1}2=7$ solutions, as indicated. Another value with many solutions is $N=5\times13\times17=1105$, namely $\frac{3\times3\times3-1}2=13$ solutions.

If one wants to disqualify solutions with a nontrivial common factor, as Jyrki Lahtonen suggests (I wouldn't know why), then each appropriate prime only contributes a factor $2$ independently of its multiplicity rather than $2m+1$ (but subtracting $1$ is omitted). This is beacuse mixing a Gaussian integer in a product with its complex conjugate introduces an integer factor, which will be common to the real and imaginary parts. In this variant one retains only $\frac{2\times2}2=2$ solutions for $N\in\{325,425\}$, and only $\frac{2\times2\times2}2=4$ solutions for $N=1105$ (namely $(73,1561)$, $(367,1519)$, $(809,1337)$, $(1057,1151)$). Even with this restriction the number will still be unbounded as $N$ acquires more and more useful prime factors.

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+1: A general formula for $n$ is probably out there. But it is a little bit trickier, when more than one conjugate pair of primes of $\mathbf{Z}[i]$ is involved, because you can allow a subset of them to be used in a way that the resulting factor is real. This shows in @Angela's list, where some pairs $(a,c)$ have a common factor 5 or 25. – Jyrki Lahtonen Dec 27 '11 at 8:57

From a beautiful theorem of Jacobi, the number of ways of writing an integer as a sum of two squares is equal to four times its number of divisors of the form $4n+1$ minus four times its number of divisors of the form $4n+3$. You can find an elementary proof of this result in Hardy & Wright, or at the end of my answer to this question if you are familiar with the basics of $L$-functions.

Hence if we choose $b$ to be, say, only divisible by primes of the form $4n+1$ then we can find arbitrarily many solutions to $2b^2=a^2+b^2$ with a fixed value of $b$.

Note: It can be easily shown that every solution of $a^2+c^2=2b^2$ in integers can be written as

$$(n^2-2mn-m^2)^2+(n^2+2mn-m^2)^2 = 2(m^2+n^2)^2$$

where $m,n$ are any integers. In particular, $b$ is itself always a sum of two squares, and we see that the number of solutions of $a^2+c^2=2b^2$ for a fixed value of $b$ is at most equal to the number of ways of writing $b$ as a sum of two squares.

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I'm somewhat confused by the reference to the result by Jacobi, which I did not know about. Using it seems all right if you are interested in counting divisors by their classes mod $4$, alternatingly, but when interested in expressions as sums of two squares, it seems rather inefficient. This is because suffices to just look at the prime factors, and only those of the form $4n+1$ (see my answer). So why would one use a result that requires enumerating all divisors instead? – Marc van Leeuwen Dec 27 '11 at 11:05
Dear Marc, I am not sure where your concerns lie. What do you mean by "inefficient"? Jacobi's theorem gives the exact number of representations as sums of two squares. It is not an algorithm, so I'm not sure whether it can be called "efficient" or "inefficient". Prime factors are one particular case of divisors, are they not? Your answer is perfectly fine and is in accordance with Jacobi's theorem. Regards, – Bruno Joyal Dec 27 '11 at 18:32
If you can describe the exact number of representations as sums of two squares either by just considering only the multiplicities of primes of the form $4n+1$ in $N$ (add one to each and multiply), or alternatively by having to count all odd divisors and doing an alternating summation, then the first method seems preferable. The number of divisors grows exponentially with the number of prime factors; also a multiplicative formula is preferable to a counting formula. I just can't see the point of using Jacobi's description unless the first one is unknown, which I cannot imagine. – Marc van Leeuwen Dec 28 '11 at 13:02