I mentioned that the number $4420$ is expressible in the form $(a^2+1)(b^2+1)$ (where $a,b$ are positive integers) in three distinct ways,here is a list of these numbers: $$4420=(1^2+1)(47^2+1)=(3^2+1)(21^2+1)=(5^2+1)(13^2+1)$$ $$26650=(2^2+1)(73^2+1)=(5^2+1)(32^2+1)=(9^2+1)(18^2+1)$$ $$……$$ $$16567585450=(13^2+1)(9872^2+1)=(47^2+1)(2738^2+1)=(278^2+1)(463^2+1)$$
I think about this problem from here:A question on elementary number theory
Here is a relative problem ：http://www.mathpages.com/home/kmath275.htm
So I wonder are there some parametric formulas for sets of these numbers(or at least for part of them)?Thanks in advance!
Here is a family of solutions: $$ m_0=-1,m_1=5,m_{k+2}=38m_{k+1}-m_k \\ c_0=1,c_1=1,c_{k+2}=38c_{k+1}-c_k \\ a_k = 4m_k+c_k+\frac{2(m_k^2+1)}{c_k} \\ b_k = a_km_k-m_k^2-1 \\ n_k = a_k^2+b_k^2 $$ So for example $$ n_1=120250=(68^2+1)(5^2+1)=(57^2+1)(6^2+1)=(43^2+1)(8^2+1)\\ n_3=505723158638933050=(98048^2+1)(7253^2+1)=(82137^2+1)(8658^2+1)=(62011^2+1)(11468^2+1) $$
To arrive at this I started by observing that $$ n = (x^2+1)(y^2+1)=(x+i)(x-i)(y+i)(y-i) $$ where $i^2=-1$. Then first multiplying $(x+i)(y+i)$ $$ n = (xy-1+(x+y)i)(xy-1-(x+y)i) = (xy-1)^2+(x+y)^2 $$ or first multiplying $(x+i)(y-i)$ $$ n = (xy+1+(y-x)i)(xy+1-(y-x)i) = (xy+1)^2+(y-x)^2 $$ so letting $b=xy-1$ we have $$ n=b^2+a^2=(b+2)^2+(a-2m)^2 $$ and given $b$ such that we can write $n$ this way we can reverse the procedure to find $x$ and $y$. Furthermore if we can write $$ n=b^2+a^2=(b+2)^2+(a-2m)^2=(b+4)^2+(a-2m_2)^2=(b+6)^2+(a-2m_3)^2 $$ then we have at least three such representations pairing $b,b+2,b+4$ with $b+2,b+4,b+6$ respectively. $$ b^2+a^2=(b+2)^2+(a-2m)^2 \implies b=am-m^2-1 \\ b^2+a^2=(b+4)^2+(a-2m_2)^2 \implies 2b=am_2-m_2^2-4 \\ b^2+a^2=(b+6)^2+(a-2m_3)^2 \implies 3b=am_3-m_3^2-9 $$ combining the first two gives $$ a(m_2-2m)=m_2^2-2m^2+2 $$ Letting $m_2=2m+1$ immediately gives a family of numbers with two representations. More generally let $m_2=2m+c$, then $$ ac = (2m+c)^2-2m^2+2 \\ a = c+4m+\frac{2m^2+2}{c} $$ Combining the expressions for $b$ and $3b$ above we get $$ m_3^2-am_3+6-3m^2+3ma=0 $$ For $m_3$ to be an integer we must have the discriminant be the square of an integer $$ d^2 = a^2-12(2+ma-m^2) $$ From here I didn't think of a good way to proceed, so using computer search for $(m,c)$ that give integers for $a$ and $d$ I found the first three solutions $(m,c)=(5,1),(191,37),(7253,1405)$ and came up with the recursion given at the top more-or-less by inspection. It remains to show that successive values for $m_k,c_k$ always result in $a_k$ and $d$ being integers, but I am confident that this is a straightforward exercise in induction and leave it to the interested reader.