A number theory problem by Aditya Narayan Sharma

Number Theory Level 3

How many 4-digit perfect squares are there such that a new 4-digit number with each of their digits increased by 1 is also a perfect square ?

1 4 None 2

1 solution

Aditya Narayan Sharma
Nov 2, 2016

If we let the number be $\displaystyle \overline{abcd}=x^2$ then according to conditions $\displaystyle \overline{\text{a+1 b+1 c+1 d+1}}=y^2$ where $x,y \in \mathbb{N},y>x$

$\displaystyle y^2=x^2+1111$

$\displaystyle (y+x)(y-x)=101.11$

Now since $x^2,y^2$ are 4-digit numbers we can say $31<x<y<100$ so that we have a rough idea that $x+y<200<101.11$ .

So in $\displaystyle (y+x)(y-x)=101.11$ the RHS is product of two distinct primes so the factors on LHS must be exactly equal to them and since $y+x>y-x$ & $y+x\ne 1111$ so the only possiblity is $\displaystyle y+x=101,y-x=11$

So we have only one number satisfying the property that is $x^2=45^2=\boxed{2025}$

Good approach. To simplify the presentation, I would advise going directly with $y^2 = x^2 + 1111$ . I don't think the first 3 lines add any value to understanding the problem. If anything, it makes the presentation more confusing.

Also, the 2nd last paragraph could be rephrased. Maybe try

$(y+x)(y-x) = 101 \times 11$ , and these terms are prime. Since $x + y < 200 \leq 101 \times 11$ , and the terms are positive integers, we must have $y+x = 101$ and $y - x = 11$ .

Calvin Lin Staff - 4 years, 7 months ago

Thanks, I've removed those three lines and added $x+y<200<101.11$

Aditya Narayan Sharma - 4 years, 7 months ago

"The number abcd is 1000a + 100b + 10c + 1d. Increasing each digit by one is equivalent to adding 1111" would be sufficient for the first part, but unless I am misreading it (and my simplified version) doesn't discuss the case when a digit is 9.

Jason Dyer Staff - 4 years, 7 months ago

A number theory problem by Aditya Narayan Sharma

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