Vtyjkyuk

I'm looking for some solutions to $varphileft(nright)=2^32$ where $varphi$ is the Euler's totient function.
I know that if $n=p_1^r_1cdotldotscdot p_k^r_k$ satisfies
$varphileft(nright)=2^32$ then
beginalign*
& 2^32=varphileft(nright)=prod_i=1^kp^r_ileft(1-frac1p_iright)=nprod_i=1^kleft(1-frac1p_iright)\
Rightarrowquad & n=frac2^32prodleft(p_i-1right)prod p_i
endalign*
So I was looking to compute solutions by finiding primes $p_i$
such that $p_i-1mid2^32$ and plug them into the last equation.
Those are $p_i-1inleft 2^lmid1leq lleq32right $ and
for example $p_i-1=2$ is good because then $p_i=3$ is a prime.
Plugging it into the equation gives
$$
n=frac2^32left(3-1right)cdot3=3cdot2^31
$$

But then $varphileft(3cdot2^31right)=varphileft(3right)varphileft(2^31right)=2left(2^31-2^30right)=2^31boldsymbolneq2^32$

What am I missing here?

In your equation for $n$ that number is also a multiple of $2$, so you must include $p_1=2$ in your product expressions. This forces tbe extra factor of $2$ into $n$ as other answers point out.

Incidentally, $3×2^32$ is not the minimal solution. The number $5×2^31$ is less, and you should experiment with other possible Fermat prime factors. Why should you use Fermat primes (along with $2$) here?

Another way to look at the totient function is
$$varphi(n)=p_1^r_1-1(p_1-1)cdot p_2^r_2-1(p_2-1)...p_k^r_k-1(p_k-1)=2^32$$
Assuming (wlog) $p_1<p_2<...<p_k$, Euclid's lemma will restrict solutions to $p_1=2$ or $r_i=1,i=overline2..k$ and $p_i=2^m_i+1$ (aka Fermat primes)(simply because if we assume $r_i>1$, then $p_i mid 2^32$ and due to Euclid's lemma this is possible for $p_i=2>p_1$). Let's see a few examples (totient function is multiplicative, this is important!)
$$varphi(2^33)=2^32$$
$$varphi(3cdot2^32)=varphi(3)cdotvarphi(2^32)=2^32$$
$$varphi(17cdot2^29)=varphi(17)cdotvarphi(2^29)=2^32$$
$$varphi(3cdot17cdot2^28)=varphi(3)cdotvarphi(17)cdotvarphi(2^28)=2^32$$
$$varphi(3cdot5cdot17cdot2^26)=varphi(3)cdotvarphi(5)cdotvarphi(17)cdotvarphi(2^26)=2^32$$
and so on ... There are $5$ Fermat primes less than $2^32$: $left 2^1+1, 2^2+1, 2^4+1, 2^8+1, 2^16+1right$. You will have to look at all the combinations $binom50+binom51+...+binom55=2^5$ and complement with some $varphi(2^n)=2^n-1$ to obtain $2^32$.