2026/062) Define $f(n)=LCM(1,2,\cdots\,n)$. Determine the smallest positive integer a such that $f(a)=f(a+2)$

This shall require that $a+1$ and $a+2$ do not increase the LCM.

This shall require both $a+1$ and $a +2 $to be composite and neither is a perfect power of a prime. 

Let us explain it for $a+1$ and the same logic holds for $s+ 2$

If $a+1$ is prime then we have not encountered the same in any of the numbers and this shall contribute to LCM and it shall increase. LCM should be multiplied by $a+ 1$ 

If  $a+1 = b^k$ where b is a prime then it should be multiplied by $b$ as $b^(k-1)$ must have come previously but not $b^k$

if $a+1=b^kc^m$ that is product of powers of 2 primes then $b^k$ and $c^m$ has already come and hence it shall not contribute to  a higher LCM

Same for $a+2$     

 So we require smallest n such that n+1, n+ 2 are composite and producut of power of at least 2 primes 

Looking at $1,2,3,4,5,6,7,8,9,10,11,12,13,14,15$

$14 = 2 * 7 $so LCM does not increase

$15 = 3 * 5$ so LCM does not increase

For $(8,9)$ both power of primes ,

 For $(9,10)$, $9$ is power of prime

 So $a = 13$ 

Next $a = 19$ as  $20=2 ^2 * 5, 21 = 3 *7$ do not increase meets criteria