Find the smallest perfect square that begins with a 1 and remains a perfect square when this 1 is replaced with a 7.
Before we get to any computer programs, there is actually a LOT of analytic work we can do.
Let the squares be m^2 and n^2. Then m^2-n^2 = 6*10^k for some integer k.
With the leading digits being 1 and 7, we can also say 2 < m/n < sqrt(7).
m^2-n^2 = (m+n) * (m-n), and this is a factorization of 6*10^k.
Both factors of 6*10^k will need to be even for an integer solution.
Then let 6*10^k factor into 4*p*q so that m+n=p and m-n=q.
Then m=p+q and n=p-q and p*q = 15*10^(k-1).
Lets plug the first two equalities into the compound inequality.
Then 2 < (p+q)/(p-q) < sqrt(7), and now some algebra:
1 < 2q/(p-q) < sqrt(7) - 1
1 > (p-q)/(2q) > (sqrt(7)+1)/6
3 > p/q > (sqrt(7)+4)/3 ~= 2.21525
This last inequality puts tight bounds on what the factorization p*q can be.
Let k=1, then p*q=15. No factorizations satisfy the inequality.
Let k=2, then p*q=150. Again, no factorizations satisfy the inequality.
Let k=3, then p*q=1500. p=60, q=25 is a valid factorization with 60/25 = 2.4.
Then m=60+25=85 and n=60-25=35. 85^2=7225 and 35^2=1225. 1225 is then the smallest number to satisfy the problem statement.
Let k=4, then p*q=15000. p=200 and q=75 is a valid factorization with 200/75 = 2.667.
Then m=200+75=275 and n=200-75=125. 275^2=75625 and 125^2=15625.
Let k=5, then p+q = 150000. p=600, q=250 is a valid factorization with 600/250 = 2.4.
Then m=600+250=850 and n=600-250=350. 850^2=722500 and 350^2=122500.
Also, p=625, q=240 is a valid factorization with 625/240 = 2.604.
Then m=625+240=865 and n=625-240=385. 865^2=748225 and 385^2=148225.
The four smallest perfect squares to satisfy the problem statement are 1225, 15625, 122500, and 148225.
Analytic Solution
