$(4t+1) (4t'+1) = (4s+1) (4s'+1)$ nonnegative integer solutions? From the book A Friendly Introduction to Number Theory by J.H.Silverman :

Welcome to $\mathbb {M}$-World, where the only numbers that exist are positive integers that leave a remainder of 1 when divided by 4.
...
We say that m $\mathbb {M}$-divides n if n = mk for some $\mathbb {M}$-number k. And we say that n is an $\mathbb {M}$-prime if its only M-divisors are 1 and itself. (Of course, we don't consider 1 itself to 
  be an M-prime.) 
...
Find an $\mathbb {M}$-number n that has two different factorizations as a product of $\mathbb {M}$-primes. 

$$$$
I tried numerically but it seems there is no example up to 101. To find a solution analytically, $(4t+1) (4t'+1) = (4s+1) (4s'+1)$ reduces to $4tt'+t+t'= 4ss'+s+s'$. I can't guess of any $s\ne t$, $s,t \in \mathbb{Z}_+$; though, how possible to (even to approach) to solve this Diyofanti equation for a general set of solutions? 
 A: There a very simple fact: a number is a $\Bbb{M}$-prime if and only if all of its prime divisors have the form $4a+3$. Now, you can consider $3 \times 3 = 9$, $7 \times 7 = 49$ and $3 \times 7 = 21$: these are $\Bbb{M}$-primes which are not prime. And magically $$441= 21 \times 21 = 9 \times 49$$ has two different factorizations.
A: I'll call a number $M$-prime if it is prime in $\mathbb M$. 
$\mathbb M$ is composed only on odd numbers, so every $n\in \mathbb M$ can be decomposed into
$$n = p_1p_2\dots p_k\cdot q_1q_2\dots q_m$$
where $p_i$ are primes in the form $4k+1$ and $q_i$ are primes in the form $4k+3$. Moreover, $n\in\mathbb M$ implies that $m$ must be even.
If $k>1$, then $n$ can be decomposed as $n= p_1 \cdot (\dots)$ where both the component are in the form $4k+1$, so $n$ is not $M$-prime. 
If $k=1$ but $m>0$ then $n$ can be decomposed as $n= p_1 \cdot (q_1q_2\dots q_m)$ so it is still not $M$-prime.
If $k=0$, but $m>2$, then $n= q_1q_2 \cdot (q_3q_4\dots q_m)$ so it is still not $M$-prime, since $q_1q_2$ and the other term  are in the form $4k+1$.
The only cases left are


*

*$k=1,m=0\implies n=p_1$

*$k=0,m=2\implies n=q_1q_2$


and they are all $M$-primes.
The only way to get a number that is product of two distinct couples of $M$-primes is to multiply two number of the second form, since
$$
a=q_1q_2\quad b=q_3q_4 \implies a\cdot b = (q_1q_2)\cdot(q_3q_4) = (q_1q_3)\cdot (q_2q_4)
$$
