STEP I 2004 question 5 solution

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STEP I 2004 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 2004 question 5 solution

$A : U_{n} = 5n - 4$

$B : U_{n} = 5n - 3$

$C : U_{n} = 5n - 2$

$D : U_{n} = 5n - 1$

$E : U_{n} = 5n$

Sum of a term in B, and a term in C:

5n - 3 + 5m - 2 = 5(n + m - 1)

This is the $(n + m - 1)^{th}$ term of E.

The square of a term in B:

$(5n - 3)^{2} = 25n^{2} - 30n + 9 = 5(5n^{2} + 6n) + 9$

$9 \equiv -1 \pmod{5}$

Hence:

$\left[ 5(5n^{2} + 6n) + 9 \right] \equiv -1 \pmod{5}$

Hence, it is expressible as:

$5(5n^{2} + 6n + 2) -1$

Which is in D.

Now consider the square of a term in C:

$(5n - 2)^{2} = 25n^{2} - 20n + 4 = 5(5n^{2} - 4n + 1) - 1$

This is again a term in D.

Assume that for some positive integers "x", and "y": $x^{2} + 5y = 24723$ .

Let "x" represent some term in B, hence $x^{2}$ is a term in D.

Hence, as $5y \equiv 0 \pmod{5}$ , $x^{2} + 5y \equiv -1 \pmod{5}$ , hence, there should be a term in D for which the equation is true, hence:

$5n - 1 = 24723 \implies 5n = 24724$

As "n" is a positive integer this is not possible for a term in B, or C (as the square is in D).

Now consider where the squares of the other sequences go:

Consider the square of a term in D:

$(5n - 1)^{2} = 25n^{2} - 10n + 1 = 5(5n^{2} - 2n + 1) - 4$ .

This is a term in A.

Consider the square of a term in E:

$(5n)^{2} = 25n^{2} = 5(5n^{2})$

This is a term in E.

Consider the square of a term in A:

$(5n - 4)^{2} = 25n^{2} - 40n + 16 = 5(5n^{2} - 8n + 4) - 4$

This is a term in A.

Hence the following can be written (note that $A^{2}$ means the square of a term in "A", etc.):

$A^{2} \to A$

$B^{2} \to D$

$C^{2} \to D$

$D^{2} \to A$

$E^{2} \to E$

There are no terms in A, or E such that the statement is true, and therefore the statement is false.

The fourth power of a term in B, or a term in C is in A.

Consider where a term in A is when two times another term in A is added to it:

5n - 4 + 10m - 8 = 5(n - 2m) - 12 = 5(n - 2m - 2) - 2 .

Hence, a term in A which has had two times another term in A added to it is in C.

Hence, consider:

$x^{4} + 2y^{4} = 26081974$ .

Now, it is possible to say that for a term in B, or C to fit the values of "x", or "y":

$5n - 2 = 26081974 \implies 5n = 26081976$

Hence there is no term in B, or C.

Now consider what happens to the other terms using the notation again:

$A^{4} \to A$

$B^{4} \to A$

$C^{4} \to A$

$D^{4} \to A$

$E^{4} \to E$

Hence, for all of the numbers that are not in E, the above argument is sufficient; now consider terms in E added together (one term added to two times another):

5n + 10m = 5(n + 2m)

This is in "E", and therefore if there existed a number for which the equation (aforementioned) was true, there would have to be (due to elimination) a term in "E" that satisfied it:

5n = 26081974 .

There is no such term.

Now consider that adding a term from E to any other term does not change the sequence it is in, hence:

$A + E \to A$

$B + E \to B$

$C + E \to C$

$D + E \to D$

$E + E \to E$ .

Now consider the possibility that "x" is from "E", and "y" is from elsewhere.

Consider 2 times a term in A:

2(5n - 4) = 10n - 8 = 2(5n - 1) - 3 .

Hence:

$2A \to B$ .

The term of $x^{4} + 2y^{4}$ would correspond to a term in E plus two times a term in A; overall a term in B; hence:

$5n - 3 = 26081974 \implies 5n = 26081977$

Hence there is no such term, and no such numbers "x", and "y".

Consider that two times a term in E is also in E, and that this plus a term in A is also in A, therefore for any integer solution to exist, the sum ( $x^{4} + 2y^{4}$ ) must be in A, hence:

$5n - 4 = 26081974 \implies 5n = 26081978$ .