STEP I 2004 5

We will let the sequence begin from n=0 for simplicity, and we’ll include 0 into sequence E…
A: a_n=1+5n
B: b_n=2+5n
C: c_n=3+5n
D: d_n=4+5n
E: e_n=5n

For any positive integer n and m
b_n+c_m=5+5(n+m)=e_{n+m}

Note: For those knowing modular arithmetic, this is exactly the same as saying that the sum of any two positive integer x,y with x \equiv 2\pmod 5 and y \equiv 3\pmod3 is divisible by 5.

For any positive integer n:

b_n^2=(2+5n)^2=4+20n+25n^2 \\ =4+5(4n+5n^2) \\ =d_{5n^2+4n}

and:

c_n^2=(3+5n)^2=9+30n+25n^2 \\ =4+5+5(6n+5n^2) \\ =4+5(5n^2+6n+1)=d_{5n^2+6n+1}
Ergo, the square of every term in C is a term in D.

Note: Again, modular arithmetic trivializes these…

i) 243723 is a term in C.

a_n^2=(1+5n)^2=1+10n+25n^2=1+5(2n+5n^2)=a_{2n+5n^2}
d_n^2=(4+5n)^2=16+40n+25n^2=1+5 \times 3+4(8n+5n^2)=a_{5n^2=8n+3}
e_n^2=5 \times 5n^2=e_{5n^2}

Hence the square of any positive integer must be in A, D, or E.

Let k \in {0,1,2,3,4} and note that k+5n+5y=k+5(n+y) so adding a multiple of 5 to any integer gives a number in the same arithmetic sequence.

Therefore, x^2+5y is in A, D, or E. Hence, we cannot have a solution as 243723 is in C.

ii) 26081974 is in D

Since the square of any integer is in A, D, or E, the fourth power must be in A or E.

2a_n=2+5(2n)=b_{2n} and 2e_n=5(2n)=e_{2n} thus 2y^4 is in B or E.

So the LHS is in E, C, A, or B, by considering all cases but never in D. Hence, there is no solution.

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