Space Bound

Number Theory Level 4

Find the number of perfect squares $\overline{ aabcd }$ with $a\neq 0$ and $d\neq 0$ such that the number $\overline { dcbaa }$ is also a perfect square.

2 4 1 3 0

1 solution

Lu Chee Ket
Oct 23, 2015

Such 5 digits of [non-zero] only valid from 100^2 to 316^2 OR 10000 to 99856:

44521 <--> 12544 {from 112^2 and 211^2}

44944 <--> 44944 {from 212^2}

no others

Since a, c and d are not restricted from being equal, 1 is not likely the answer wanted, 2 is the answer.

we accept from 100 to 316, let (m 100+n 10+p)^2=dbcaa,,,, n[0->9], m[1->3], p[1-9]

then 10000m^2+100n^2+p^2+2000mn+200mp+20np=dbcaa

let consider only p^2+20np that determine aa or a*10+a

if p=1, a=1, but 20np always even then cannot be 1

p=2, a=4, we find that n must be 1 or 6 so that aa can be 44

p=3, just like p=1 not eligible

p=4, a=6, p^2 =16 then the 10*a (the ten place) will be odd (20np + 10 (from p^2))

p=5,6,7,9 also not.

p=8 eligible that a also =4, 160n+64 then n can be 3 or 8

then it should be 44bcd and dcb44

and we know that (m 100+n 10+p)^2=dbc44

Now the work more easily with m from 1 to 3, n can be 1, 6 for p=2, n can be 3 or 8 for p=8.

Just subs those value to find dbc44, then reverse it and square root it for finding the right answers.

HIEU TRAN - 5 years, 7 months ago

A much simpler way is to show that a perfect square mod 4 = 0, 1 only. So $a = 4,7,8$ only. But because no values of $x^2$ ends in $7$ or $8$ for integer $x$ , then $a = 7$ and $a=8$ can't be a solution. Then $a = 4$ only. Can you take it from here?

Pi Han Goh - 5 years, 7 months ago

One (very slight) omission: you didn't explain the elimination of a=3. Otherwise, I completely agree with your way of doing it.

Peter Byers - 5 years, 7 months ago

@Peter Byers – Oh right thanks! I wasn't looking when I was typing. haha

Pi Han Goh - 5 years, 7 months ago

Space Bound

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