Impossible Exponential Diophantine Equation

Diophantine equations that mix exponentials and polynomials can seem intimidating, but the key to proving no integer solution exists often lies in a clever modular reduction. In this post, we work through an HSC Maths Extension 2–style problem, using the Binomial Theorem and congruences to narrow down the possibilities to just a handful of values, then checking each. Along the way, you'll see how a single modular insight can turn an impossible-looking equation into a routine verification.

Problem Statement

Explain why there is no integer $n$ such that $(n+1)^{41} - 79n^{40} = 2$.

---

Hints

Attempt the proof independently first. Focus on the key theorem, algebraic transformation, or contradiction setup that links the hypothesis to the target conclusion.

---

Solutions

Proof by Modular Analysis and Case Checking

Before diving into heavy algebra, we quickly test the simplest integer — zero — just to rule it out. After that, we'll move to a more powerful modular technique for non‑zero $n$.

Step 1: Test $n = 0$

If $n = 0$:

$$(0+1)^{41} - 79(0)^{40} = 1 - 0 = 1 \neq 2$$

So $n = 0$ is not a solution.

Step 2: Analyze for $n \neq 0$ using modular arithmetic

Now that we know $n$ is non‑zero, we can safely work modulo $n$. This will allow us to cancel large powers of $n$ and leave a very simple congruence.

For $n \neq 0$, consider the equation modulo $n$:

By the Binomial Theorem:

$$(n+1)^{41} = \sum_{k=0}^{41} \binom{41}{k} n^k \cdot 1^{41-k}$$

All terms contain $n$ except the last term ($k=0$):

$$(n+1)^{41} \equiv 1 \pmod{n}$$

The term $-79n^{40}$ is clearly divisible by $n$:

$$-79n^{40} \equiv 0 \pmod{n}$$

Therefore:

$$\begin{aligned} (n+1)^{41} - 79n^{40} &\equiv 2 \pmod{n} \\ 1 - 0 &\equiv 2 \pmod{n} \\ 1 &\equiv 2 \pmod{n} \end{aligned}$$

This congruence points to a very limited set of possibilities for $n$.

Step 3: Interpret the congruence

$1 \equiv 2 \pmod{n}$ means $n \mid (1-2)$, so $n \mid (-1)$.

Therefore $n \in \{-1, 1\}$.

Step 4: Test candidate values

With only two candidates left, we simply substitute them back into the original equation to verify.

For $n = 1$:

$$(1+1)^{41} - 79(1)^{40} = 2^{41} - 79 = 2,199,023,255,552 - 79 \neq 2$$

For $n = -1$:

$$(-1+1)^{41} - 79(-1)^{40} = 0 - 79(1) = -79 \neq 2$$

Conclusion

All possible integer values of $n$ have been checked and none satisfy the equation.

Therefore, there is no integer $n$ such that $(n+1)^{41} - 79n^{40} = 2$.

---

Takeaways

• Modular Arithmetic Strategy: Working mod $n$ can dramatically constrain possible solutions
• Binomial Expansion Mod $n$: $(n+1)^k \equiv 1 \pmod{n}$ since all terms except constant contain $n$
• Divisor Analysis: $1 \equiv 2 \pmod{n}$ means $n \mid (-1)$, giving only $n \in \{-1, 1\}$
• Exhaustive Checking: After constraining to finitely many cases, verify each directly

---

Further Readings

If you found this proof interesting, be sure to check out these relevant HSC booklets to sharpen your reasoning skills: HSC Proofs, HSC Vectors, HSC Last Resorts