A350870 Numbers k = x.y such that x.y = (x+y)^2, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

81, 2025, 3025, 494209, 24502500, 25502500, 52881984, 60481729, 6049417284, 6832014336, 101558217124, 108878221089, 123448227904, 127194229449, 152344237969, 213018248521, 217930248900, 249500250000, 250500250000, 284270248900, 289940248521, 371718237969, 413908229449, 420744227904

Offset: 1

Author

Bernard Schott, Jan 20 2022

Keywords

• nonn
• base

Comments

Problem proposed on French site Diophante (see link).

We have to solve Diophantine equation (x+y)^2 = x*10^m + y where m = length(x) = length(y).

Squares of a variant of Kaprekar numbers (A045913).

Number of solutions with 2*m digits for m >= 1: 1, 2, 1, 4, 2, 21, ...

Compare with A347541 where x*y divides x.y, when x and y have the same number of digits, "." means concatenation, and y may not begin with 0.

Examples

			81 = (8+1)^2, hence 81 is a term.
3025 = (30+25)^2, hence 3025 is another term.
		

Links

• Diophante, A1945 - Concaténations en tous genres (in French).

Crossrefs

Equals A238237 \ A350918.

Subsequence of A102766.

Subsequences: A038544, A350869.

Cf. A045913, A347541.

Programs

PARI

upto(n) = {i = 4; i2 = i^2; res = List(); while(i2 <= n, i++; i2 = i^2; if(#digits(i2) % 2 == 1, i = sqrtint(10^(#digits(i2))) + 1; i2 = i^2; ); if(is(i2), listput(res, i2) ); ); res }
is(n) = { my(d = digits(n), d1, d2, frd2); if(#d % 2 == 1, return(0)); d1 = vector(#d \ 2, i, d[i]); d2 = vector(#d \ 2, i, d[i + #d \ 2]); frd2 = fromdigits(d2); 10^(#d \ 2 - 1) <= frd2 && (fromdigits(d1) + frd2)^2 == n } \\ David A. Corneth, Jan 21 2022

Formula

a(n) = A045913(n+1)^2.