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If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: $$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$ In fact, for $0\leq i\leq k\geq2$, we have $$\text{gcd}(a+i,b-i)=i+2.$$

BTW, very cool drawing, Joseph!

If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: for $0\leq i\leq k\geq2$, we have $$\text{gcd}(a+i,b-i)=i+2>1.$$

BTW, very cool drawing, Joseph!

If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: $$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$

BTW, very cool drawing, Joseph!

If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: $$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$ In fact, for $0\leq i\leq k\geq2$, we have $$\text{gcd}(a+i,b-i)=i+2.$$

BTW, very cool drawing, Joseph!

If you're not into "hunting for primes", here is a slight variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: $$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$

If you're not into "hunting for primes", here is a slight and concrete variant.

Let $a-2=(k+2)!$ and $b+2=(k+2)!$. Then proceed in the same manner as Fedor's example: $$\text{gcd}(a+i,b-i)=\text{gcd}((a-2)+i+2,b+2-(i+2))>1.$$

BTW, very cool drawing, Joseph!