# Arithmetic Progression proof for sum of an integer

Question:-

The fourth power of the common difference of an arithmetic progression with integer entries is added to the product of any four consecutive terms of it. Prove that resulting sum is the square of an integer.

MyApproach:

Let d be the common difference and d^4 be the fourth power.I am considering the four consecutive terms be (a+2d),(a+d),(a-d),(a-2d).What do the question mean by telling that the common difference is added to the four consecutive terms ?What is the product of the terms equals to ?

• First, you haven't identified four consecutive terms: you skipped right over the middle term $a$. So it should be adjusted to (for example) $a - d, a, a + d, a + 2d$. The product of four consecutive terms is $(a + 2d)(a + d)a(a - 2d)$. – user296602 Feb 11 '18 at 16:11

If the common difference is $2d$,

$$(a-3d)(a-d)(a+d)(a+3d)+(2d)^4$$

$$=(a^2-d^2)(a^2-9d^2)+(2d)^4$$

$$=(a^2)^2-2a^2(5d^2)+(5d^2)^2$$

$$=?$$

• Is it (a^2-(5d^2))^2? Is this the answer that is asked. – user517784 Feb 11 '18 at 16:14
• @CalculusProgrammer, $$(x-y)^2=?$$ – lab bhattacharjee Feb 11 '18 at 16:15
• Yes thanks for showing the steps. – user517784 Feb 11 '18 at 16:18