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edited Feb 11, 2022 at 10:09

Python 2, 68 bytes

Inefficient and somewhat boring approach.

lambda n:`n`in('%d'*2%(a*a,b*b)for a in range(1,n)for b in range(n))

Much more efficient and interesting.

f=lambda n,d=10m=0,d=1:n/d and((d/11+n%10>0<d>n%d*10)==(n%dm*10<d!=10)**==n**.5%1+(n/d)**5%1+m**.5%1)|f(n/10,m+n%10*d,d*10)

Inefficient and somewhat boring approach.

lambda n:`n`in('%d'*2%(a*a,b*b)for a in range(1,n)for b in range(n))

Much more efficient and interesting.

f=lambda n,d=10:n/d and((d/11+n%10>0<d>n%d*10)==(n%d)**.5%1+(n/d)**.5%1)|f(n,d*10)

Inefficient and somewhat boring approach.

lambda n:`n`in('%d'*2%(a*a,b*b)for a in range(1,n)for b in range(n))

Much more efficient and interesting.

f=lambda n,m=0,d=1:n and((m*10<d!=10)==n**.5%1+m**.5%1)|f(n/10,m+n%10*d,d*10)

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created Feb 11, 2022 at 9:57

Inefficient and somewhat boring approach.

lambda n:`n`in('%d'*2%(a*a,b*b)for a in range(1,n)for b in range(n))

Much more efficient and interesting.

f=lambda n,d=10:n/d and((d/11+n%10>0<d>n%d*10)==(n%d)**.5%1+(n/d)**.5%1)|f(n,d*10)