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Let the seven squares be $a_1^2,a_2^2,\dots ,a_7^2$ where $a_i$ are positive integers.

Consider the possible residues of $a^2$ modulo $20$.

Since $a$ can be any integer, $a^2$ modulo $20$ depends on $a$ modulo $20$.

Let us compute $a^2\mathrm{mod}20$ for $a=0,1,\dots ,19$:

$a$	$a^2\mathrm{mod}20$
0	0
1	1
2	4
3	9
4	16
5	5
6	16
7	9
8	4
9	1
10	0
11	1
12	4
13	9
14	16
15	5
16	16
17	9
18	4
19	1

So the possible residues are $0,1,4,5,9,16$.

Counting the number of distinct residues: - $0$ (for $a\equiv 0,10(\mathrm{mod}20)$) - $1$ (for $a\equiv 1,9,11,19(\mathrm{mod}20)$) - $4$ (for $a\equiv 2,8,12,18(\mathrm{mod}20)$) - $5$ (for $a\equiv 5,15(\mathrm{mod}20)$) - $9$ (for $a\equiv 3,7,13,17(\mathrm{mod}20)$) - $16$ (for $a\equiv 4,6,14,16(\mathrm{mod}20)$)

There are only $6$ possible residues for squares modulo $20$.

By the pigeonhole principle, among any $7$ squares, at least two must have the same residue modulo $20$.

Therefore, their difference is divisible by $20$.